NCERT Exemplar Class 12 Maths Differential Equations - Class 1...

Pattern-recognition angle. Whenever the RHS is a ^{linear in x,y}, split the exponent into an x-part and a y-part using a^u+v=a^ua^v. After that the equation is always separable.

1. Identify the form: dydx=2^y· 2^-x, a product of a function of y and a function of x.
2. Cross-multiply: 2^-y dy = 2^-x dx.
3. Integrate term-by-term: -2^-yln 2 = -2^-xln 2 + C₁.
4. Tidy: 2^-y=2^-x+C where C=-C₁ln 2 is an arbitrary constant.
5. Sanity check by implicit differentiation: -2^-yln 2· y' = -2^-xln 2, i.e. y'=2^x-y· 2^-2x· 2^x=2^y-x.

Why this matters. Recognising the a^u+v split is a recurring trick in solving exponential separable DEs (e.g. Q61 of this Exemplar uses the same idea).

2^-y=2^-x+C.

Geometric angle. A non-vertical line has constant slope. The geometric condition ``slope is constant'' translates directly into ``the second derivative is zero''.

1. Geometric fact: along any straight line y'=m=const.
2. Differentiating the constant gives y''=0.
3. Conversely, integrating y''=0 twice produces y=mx+c, recovering every non-vertical line.

Why this matters. The same logic shows ``all non-horizontal lines x=my+c'' satisfy d²xdy²=0 (see Q77(xi)).

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

y''=0.

Definite-integral angle. Instead of finding the constant from an initial condition, integrate as a definite integral from the known point.

1. Separate: e^2y dy=dx.
2. Definite-integrate from the known point (x,y)=(5,0) to the unknown (x,3): ₀³ e^2y dy = ₅^x dt.
3. Evaluate the LHS: [e^2y2]₀³ = e⁶-12. RHS: x-5.
4. Solve: e⁶-12=x-5 x=5+e⁶-12=e⁶+92.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

x=e⁶+92.

Recognise-the-product angle. Notice that (x²-1) y' + 2xy = ddx[(x²-1) y] already, by the product rule, before we ever write the integrating factor.

1. Compute ddx[(x²-1) y] = (x²-1) y'+ 2xy. This is exactly the LHS of the given equation.
2. So the DE becomes ddx[(x²-1) y]=1x²-1.
3. Integrate using 1x²-1=12(1x-1-1x+1): (x²-1) y = 12(ln|x-1|-ln|x+1|)+C = 12ln|x-1x+1|+C.

Why this matters. Spotting a ready-made product-rule LHS bypasses the I.F. computation entirely; the I.F. method just forces the LHS into this form.

y(x²-1)=12ln|x-1x+1|+C.

I.F. angle. The equation is also linear: y'+(2x-1) y=0, so we can use I.F.=e^{∫(2x-1) dx}=e^x2-x.

1. Multiply through by I.F. e^x2-x: ddx[y e^x2-x] = 0.
2. Integrate: y e^x2-x=C, hence y=C e^-(x2-x)=C e ^x-x2.
3. Cross-check by differentiation: y'=C e^x-x2(1-2x)=y(1-2x), which matches the given DE.

y=C e ^x-x2.

Particular + homogeneous angle. For a linear DE with constant coefficient, write y=y_p+y_h: a particular solution of the inhomogeneous equation plus the general solution of the homogeneous one.

1. Homogeneous part: y'+ay=0 ⇒ y_h=C e^-ax.
2. Try y_p=A e^mx. Substituting: Am e^mx+aA e^mx=e^mx, so A(m+a)=1, giving A=1m+a (assuming m+a≠ 0).
3. Combine: y=e^mxm+a+C e^-ax.
4. When m=-a, the trial A e^mx duplicates the homogeneous solution and the rule is to multiply by x: try y_p=A x e^-ax; substituting gives A=1, hence y=(x+C) e^-ax.

Why this matters. The same particular-plus-homogeneous decomposition underlies every constant-coefficient linear DE later in higher mathematics.

y=e^mxm+a+C e^-ax (or (x+C)e^-ax if m=-a).

Direct angle. The equation dydx=e^x+y-1 already has the special structure f(x+y) on the RHS, which is the textbook signal for v=x+y.

1. With v=x+y and v'=1+y', the DE becomes v'-1=e^v-1, i.e. dvdx=e^v.
2. Cross-multiply and integrate: -e^-v=x+C.
3. Replace v and tidy: x+e^-(x+y)+C=0, i.e. e^-(x+y)=-(x+C).

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

e^-(x+y)+x+C=0.

Pull out the differential of y/x. The quantity y dx-x dyx² is exactly -d(yx). That suggests dividing the given equation by x².

1. Divide y dx-x dy=x²y dx by x²: y dx-x dyx² = y dx -d(yx) = y dx.
2. Let u=yx. Then y=u x and y dx=u x dx, while -du=u x dx, i.e. duu=-x dx.
3. Integrate: ln|u|=-x²2+C₁, so u=C e^-x2/2, i.e. yx=C e^-x2/2.

Why this matters. Recognising y dx-x dy=-x² d(y/x) is a small but high-leverage trick that unlocks many ``mixed dx,dy'' problems.

y=C x e^-x2/2.

Spot-the-factoring angle. Whenever the RHS is a polynomial in x and y², try to write it as f(x)· g(y). Common factorisations: 1+x+y²+xy²=(1+x)(1+y²); x+y+xy+1=(1+x)(1+y); etc.

1. Group: 1+x+y²+xy²=(1+x)(1+y²).
2. Use the separable template ∫ g(y) dy=∫ f(x) dx: ∫ dy1+y²=∫ (1+x) dx.
3. Get tan^-1y=x+x²2+C and use y(0)=0 to set C=0.

Why this matters. The factoring habit will resurface in Q15, Q23 of this Exemplar.

y=tan(x+x²2).

Treat x as the dependent variable. Whenever the unknown x appears linearly but y appears in a complicated way, swap roles.

1. Form: dxdy-1y x=2y², linear in x.
2. I.F.=e^-ln|y|=1/y.
3. Standard formula: x.F.=∫ Q.F. dy+C gives xy=∫ 2y dy=y²+C.
4. Multiply by y: x=y³+Cy.
5. Cross-check by direct substitution: dydx=1dx/dy=13y²+C; and yx+2y³=yy³+Cy+2y³=y3y³+Cy=13y²+C. Match.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

x=y³+Cy.

First-integral angle. Multiply through by 1+y and notice that (2+sin x) dy+(1+y)cos x dx=0 is the differential of the product (1+y)(2+sin x).

1. Rearrange to (2+sin x) dy + (1+y)cos x dx = 0.
2. Observe d[(1+y)(2+sin x)]=(2+sin x) dy+(1+y)cos x dx. Hence d[(1+y)(2+sin x)]=0.
3. Integrate: (1+y)(2+sin x)=C. Use y(0)=1 to get C=4.
4. At x=π/2: (1+y)(3)=4 ⇒ y=1/3.

Why this matters. ``Spot the exact differential'' is the fastest method for separable problems with a product structure.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

y(π/2)=1/3.

Closed-form angle. Solve for y(t) in closed form, then evaluate at t=1.

1. With I.F.=(1+t)e^-t, the standard formula gives y(1+t)e^-t=∫ e^-t dt + C = -e^-t+C.
2. So y=-e^-t+C(1+t)e^-t=-1+C e^t1+t.
3. Apply y(0)=-1: -1+C1=-1 ⇒ C=0.
4. Hence y(t)=-11+t. At t=1: y(1)=-12.

Why this matters. Whenever the question asks for a specific function value, finding a closed form first is usually easier than juggling definite integrals.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

y(t)=-11+t, so y(1)=-12.

Trig substitution angle. Put x=sinθ; then √1-x²=cosθ and cos^-1x=π2-θ.

1. y=θ²+A(π2-θ)+B=θ²-Aθ + (Aπ2+B).
2. Differentiate w.r.t. x using dθdx=1cosθ: dydx=(2θ-A)·1√1-x².
3. So √1-x² y'=2θ-A; differentiating once more leads to (1-x²)y''-xy'=2 exactly as in the main solution.

Why this matters. The substitution x=sinθ neutralises every √1-x² in sight.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

(1-x²)y''-xy'=2.

Direct elimination. Use the implicit derivative and the original equation simultaneously to eliminate a.

1. Implicit derivative of x²+y²-2ay=0: 2x+2yy'-2ay'=0 ⇒ a=x+yy'y'.
2. From original: a=x²+y²2y.
3. Equate: x+yy'y'=x²+y²2y.
4. Cross-multiply: 2y(x+yy')=(x²+y²)y' ⇒ (x²-y²)y'=2xy.

Why this matters. The same elimination pattern appears in MCQ Q59 of this Exemplar.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

(x²-y²)y'=2xy.

Standard I.F. angle.

1. Divide by 1+x²: y'+2x1+x² y=4x²1+x².
2. I.F.=e^{∫2x1+x2 dx}=e^ln(1+x2)=1+x².
3. (1+x²) y=∫(1+x²)·4x²1+x² dx=4x³3+C.
4. C=0 from y(0)=0, so y=4x³3(1+x²).

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

y=4x³3(1+x²).

Recognise-homogeneous angle. Total degree 2 in every term ⇒ homogeneous.

1. Every term (x²,xy,y²) has degree 2, so the equation is homogeneous.
2. y=vx gives xdvdx=1+v², a separable equation.
3. Integrate: tan^-1v=ln|x|+C.
4. Restore: tan^-1(y/x)=ln|x|+C.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

tan^-1(y/x)=ln|x|+C.

Substitution-first angle. Set u=tan^-1y. Then du=dy1+y², and the linear DE becomes dxdu+x=e^u.

1. I.F. of the simplified DE is e^u.
2. ddu(x e^u)=e^2u, so x e^u=e^2u2+C.
3. Restore u=tan^-1y: 2 x e^{arctan y}=e^{2arctan y}+C'.

Why this matters. Substituting away an awkward inverse function before applying I.F. is a recurring time-saver.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

2x e^{arctan y}=e^{2arctan y}+C.

Choose-the-cleaner-variable angle.

1. Same substitution x=vy.
2. Reduces to (v²+1) dy+y dv=0.
3. Integrates to tan^-1v=-ln|y|+C.
4. Restore: tan^-1(x/y)+ln|y|=C.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

tan^-1(x/y)+ln|y|=C.

Hint-driven angle.

1. Substitute z=x+y; then z'-1=z-1z+1, so z'=2zz+1.
2. Separate: (12+12z) dz=dx.
3. Integrate: z2+ln|z|2=x+C₁.
4. Restore z and tidy: ln|x+y|=x-y+C.

Why this matters. Any RHS of the form F(x+y) yields to z=x+y.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

ln|x+y|=x-y+C.

Logarithmic-combination angle.

1. Separation (1-3y+3)dy=2xdx integrates to y-3ln|y+3|=2ln|x|+C₁.
2. Use y(1)=-2: C₁=-2.
3. Combine logs: y+2=ln|x|²+ln|y+3|³=ln(x²|y+3|³).
4. Exponentiate: x²(y+3)³=e^y+2.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

x²(y+3)³=e ^y+2.

Polish the answer. The final form ysin x=3-cos 2x2 simplifies using cos 2x = 1-2sin²x.

1. Linear-DE machinery (see main solution) gives ysin x=-cos 2x2+C.
2. Initial condition pins C=32, so ysin x=3-cos 2x2.
3. Substitute cos 2x = 1-2sin²x: 3-(1-2sin²x)2=2+2sin²x2=1+sin²x.
4. Hence y=1+sin²xsin x=csc x+sin x.

Why this matters. Always simplify the answer using standard trig identities; the marker rewards the cleanest form.

y=csc x+sin x.

Bookkeeping angle. Use (*) to express A/B, then substitute into the equation obtained by differentiating (*) again.

1. From (*): AB=-y y'x. Divide (**) by B: AB+(y')²+y y''=0.
2. Substitute: -y y'x+(y')²+y y''=0.
3. Multiply by x: -y y'+x(y')²+xy y''=0, i.e. xy y''+x(y')²-y y'=0.
4. Verify at A=B=1 (a unit circle x²+y²=1): y'=-x/y, y''=-1/y - x· y'/y²=-1/y - x(-x/y)/y²=-1/y³·(y²+x²)=-1/y³ (using x²+y²=1). Plug in and confirm the equation holds.

xy y''+x(y')²-y y'=0.

Both pieces are differentials. Observe both terms are exact differentials of simple expressions.

1. tan^-1x1+x² dx = d((tan^-1x)²2).
2. 2y1+y² dy = d(ln(1+y²)).
3. Their sum being zero means d((tan^-1x)²2+ln(1+y²))=0.
4. Integrate: (tan^-1x)²+2ln(1+y²)=C.

Why this matters. Recognising exact differentials lets you skip the substitution step entirely.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

(tan^-1x)²+2ln(1+y²)=C.

Geometric angle. The tangent to a circle is perpendicular to the radius at the point of tangency.

1. Slope of radius from (1,2) to (x,y): y-2x-1.
2. Perpendicularity condition: dydx·y-2x-1=-1.
3. Cross-multiply: (y-2) dydx=-(x-1), i.e. (x-1)+(y-2)dydx=0.

Why this matters. The same ``radius ⊥ tangent'' argument generates the DE of any one-parameter family of circles with a fixed centre.

(x-1)+(y-2) y'=0.

Two-integral-then-combine angle. Compute each integral on the RHS separately first; assemble at the end.

1. Set I₁=∫ x²sin x dx, I₂=∫ x²ln x dx.
2. Tabular by-parts for I₁: differentiate x²→ 2x→ 2→ 0; integrate sin x→ -cos x→ -sin xx. Then I₁=x²(-cos x)-2x(-sin x)+2(cos x)=-x²cos x+2xsin x+2cos x.
3. For I₂ with u=ln x, du=dxx and dv=x²dx, v=x³/3: I₂=x³ln x3-∫ x²3 dx=x³ln x3-x³9.
4. Assemble: x²y=I₁+I₂+C. Divide by x².
5. The -cos x in -x²cos x divided by x² gives -cos x; the rest become xln x3-x9 etc.

Why this matters. Tabular by-parts cuts the bookkeeping in half for any ∫ xⁿsin x dx or ∫ xⁿcos x dx.

y=xln x3-x9-cos x+2sin xx+2cos xx²+Cx².

Pre-multiply by cos y. Rewriting 1+tan y=cos y+sin ycos y removes the trig fraction and makes the I.F. cleaner.

1. Multiply the original DE by cos y: (cos y+sin y)(dx-dy)+2xcos y dy=0.
2. Rearrange: (cos y+sin y) dx = [(cos y+sin y)-2xcos y] dy.
3. Divide by cos y+sin y: dxdy+2cos ycos y+sin y x = 1.
4. Computing ∫ P₁ dy=y+ln|sin y+cos y|, I.F. =e^y(sin y+cos y).
5. Standard linear-DE machinery gives x(sin y+cos y)=sin y+C e^-y.

Why this matters. Pre-clearing trig fractions before applying I.F. keeps the integrals tractable.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

x(sin y+cos y)=sin y+C e^-y.

Half-angle simplification. The identities 1+cos z=2cos²(z/2) and sin z=2sin(z/2)cos(z/2) unlock the separation.

1. With z=x+y: dzdx=1+cos z+sin z=2cos(z/2)[cos(z/2)+sin(z/2)].
2. Divide RHS by cos²(z/2): 2sec(z/2)[sec(z/2)+tan(z/2)sec(z/2)]²(z/2)=2cos(z/2)[cos(z/2)+sin(z/2)]. Equivalently dz1+cos z+sin z=dx, and the substitution u=1+tan(z/2) collapses the LHS.
3. Integrate: ln|1+tan(z/2)|=x+C₁, hence 1+tan(x+y2)=Ce^x.

Why this matters. The Weierstrass-type substitution t=tan(z/2) underlies many trigonometric DEs.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

1+tan(x+y2)=Ce^x.

Undetermined-coefficients angle. Try a particular solution of the form y_p=Asin 2x+Bcos 2x.

1. Homogeneous: y_h=Ce^3x (from y'-3y=0).
2. Try y_p=Asin 2x+Bcos 2x. Then y_p'=2Acos 2x-2Bsin 2x.
3. Substitute into y'-3y=sin 2x: (2Acos 2x-2Bsin 2x) - 3(Asin 2x+Bcos 2x) = sin 2x. Equating coefficients: cos 2x: 2A-3B=0; sin 2x: -2B-3A=1.
4. Solve: from first, A=3B2. Substituting: -2B-9B2=1 ⇒ -13B2=1 ⇒ B=-213, A=-313.
5. So y_p=-3sin 2x+2cos 2x13 and the general solution is y=y_p+y_h.

Why this matters. Undetermined coefficients is faster than the formula whenever the forcing term is e^axsin bx, e^axcos bx or a polynomial in x.

y=-3sin 2x+2cos 2x13+C e^3x.

Sanity-check angle. Cross-check the answer by plugging (2,1) in: 2(4-1)=6=3· 2.

1. Homogeneous DE dydx=1+(y/x)²2(y/x), substitution y=vx gives x v'=1-v²2v.
2. 2v dv1-v²=dxx integrates to -ln|1-v²|=ln|x|+C₁.
3. Exponentiate: 11-v²=Cx, so x²-y²=kx.
4. Initial condition (2,1): 3=2k⇒ k=3/2, giving 2(x²-y²)=3x.

Why this matters. The locus turns out to be a circle: 2x²-2y²-3x=0 rearranges to (x-3/4)²-y²=9/16, a rectangular hyperbola through (2,1).

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

2(x²-y²)=3x.

Partial-fractions angle.

1. 1x(x+1)=1x-1x+1 (cover-up).
2. Integrate: ln|y-1|=ln|xx+1|+C₁, so y-1=C xx+1.
3. (1,0)⇒ C=-2, giving y=1-x1+x.
4. Check: at x=1, y=0. At x=0, y=1. Derivative gives (-1)(1+x)-(1-x)(1+x)²=-2(1+x)², and y-1x(x+1)=-2x/(1+x)x(x+1)=-2(1+x)².

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

y=1-x1+x.

Hyperbolic-tan angle. Recognise that ∫ dv1-v²=tanh^-1v (for |v|<1), which gives a tidier closed form.

1. DE becomes dv1-v²=dx with v=x-y.
2. tanh^-1v = x+C₁. At origin v=0, C₁=0.
3. So tanh^-1(x-y)=x, i.e. x-y=tanh x, equivalently y=x-tanh x.
4. Cross-check at origin: y(0)=0. Slope: y'=1-sech²x=tanh²x=(x-y)².

Why this matters. The two answer forms 1+x-y1-x+y=e^2x and x-y=tanh x are equivalent (use tanh x=e^2x-1e^2x+1).

x-y=tanh x, i.e. 1+x-y1-x+y=e^2x.

Intercept-form angle. The mid-point condition is equivalent to A=(2x,0) and B=(0,2y).

1. For P to be the mid-point of A=(a,0) and B=(0,b), we need a=2x and b=2y.
2. Intercept form of the tangent: X2x+Y2y=1. At P=(x,y): 12+12=1.
3. Slope of this line: -2y2x=-yx. Hence dydx=-yx.
4. Integrate to xy=C. Initial condition (1,1) gives C=1, so xy=1.

Why this matters. Recognising the geometric condition's intercept-form equivalent skips three lines of algebra.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

xy=1.

Nested-substitution angle. Two substitutions (v=y/x, then u=log v) reduce the DE to the simplest form duu=dxx.

1. Homogeneous: y=vx ⇒ x v'=vlog v.
2. Inner substitution u=log v: v'=e^u u' and v=e^u, so x e^u u' = e^u· u, i.e. x u' = u.
3. Separate: duu=dxx, integrate to ln|u|=ln|x|+C₁, hence u=Cx.
4. Restore: log(y/x)=Cx, i.e. y=x e^Cx.

Why this matters. Spotting that dvvlog v=d(loglog v) is a near-Olympiad trick worth remembering.

y=x e^Cx.

Quick reading. Scan for transcendental functions wrapped around any derivative. One hit ⇒ degree undefined.

1. Spot sin(dydx) on the RHS.
2. Spot the rule: transcendental function of a derivative ⇒ degree undefined.
3. Eliminate (A), (B), (C). Choose (D).
4. Order is still defined: it is 2 (from the d²ydx² term).

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

(D).

Polynomial-form check.

1. Expand [1+(y')²]³ as a polynomial in y': 1+3(y')²+3(y')⁴+(y')⁶.
2. Now the equation is 1+3(y')²+3(y')⁴+(y')⁶=(y'')², polynomial in both y' and y''.
3. Highest-order derivative is y''; its power in the equation is 2.

Numerical sanity check. Substitute a simple specific value (or limit) into both the original DE and the integrated form to confirm consistency. Even a one-point check rules out the most common algebraic slips (sign errors, dropped factors of two, missing +C).

Reasoning recap. The strategy of identifying the standard form first (separable / linear / homogeneous), then applying the matching template, is the single biggest time-saver in any differential-equations exam. Marking schemes typically award separate credit for: (i) correct identification of the form, (ii) correct intermediate algebra, (iii) the final answer with an arbitrary constant present where required. Show each of these clearly.

(D).

Cross-check by isolating the fractional term.

1. Rewrite: (dydx)^1/4=-d²ydx²-x^1/5.
2. Raise both sides to the 4th power to attempt to clear the 1/4: dydx=(d²ydx²+x^1/5)⁴. Expanding this binomial keeps mixing d²ydx² with x^1/5 (an x-only term) — fine for x but the original fractional power on dydx has now propagated into a higher-degree mix with x^1/5.
3. There is no finite polynomial form simultaneously free of fractional powers of both dydx and d²ydx², so the strict polynomial-in-derivatives form does not exist.
4. Conclude: order 2, degree undefined.

Common alternative. A second valid route is to differentiate the candidate answer and confirm it satisfies the original DE — this catches sign errors that ``forward'' integration sometimes hides. For any boxed answer F(x,y)=C, differentiate implicitly and rearrange; you must recover the given DE.

Marking-scheme angle. Examiners look for (i) the standard-form identification, (ii) the correct intermediate work, and (iii) the boxed final answer with an arbitrary constant. Skipping any of the three risks dropped marks even when the final answer is correct.

(A).