Differential Equations Class 12 NCERT Solutions Chapter 9 Ex 9.2

Concept used. A function y=φ(x) is a solution of a differential equation if substituting φ (and its derivatives) into the equation yields a true identity in x. We compute the necessary derivatives and substitute.

1. Start with y = e^x+1. Differentiating once with respect to x: y' = ddx(e^x+1) = e^x+0 = e^x.
2. Differentiating again: y'' = ddx(e^x) = e^x.
3. Substitute into the LHS of the DE y''-y': y''-y' = e^x-e^x = 0.
4. The RHS is 0. So LHS = RHS for every x∈R.

Hence y = e^x+1 is a solution of y''-y'=0.

Concept used. Substitute the given function and its derivative into the DE; check the identity.

1. Differentiate y = x²+2x+C once with respect to x. Since ddx(C)=0, y' = 2x+2.
2. Plug into the LHS of the DE: y'-2x-2 = (2x+2)-2x-2 = 0.
3. LHS = 0 = RHS for every x∈R.

Hence y = x²+2x+C is a solution.

Concept used. Differentiate and substitute into the DE.

1. y = cos x + C y' = -sin x + 0 = -sin x.
2. Substitute into LHS of the DE: y' + sin x = -sin x + sin x = 0.
3. LHS = 0 = RHS for every x∈R.

Hence y = cos x + C is a solution.

Concept used. Use the chain rule on √1+x² = (1+x²)^1/2, then verify the algebraic identity.

1. Differentiate y = (1+x²)^1/2 using the chain rule: y' = 12(1+x²)^-1/2· ddx(1+x²) = 12(1+x²)^-1/2· 2x.
2. Simplify: y' = x√1+x².
3. Express the RHS of the DE in terms of x alone using y=√1+x²: xy1+x² = x√1+x²1+x² = x√1+x².
4. LHS = RHS. Identity verified for all x∈R.

Hence y = √1+x² is a solution.

Concept used. Substitute the function and its derivative into the DE.

1. y = Ax y' = A.
2. Plug into the LHS of the DE: xy' = x· A = Ax.
3. The RHS is y = Ax.
4. LHS = Ax = RHS for all x≠ 0.

Hence y = Ax is a solution.

Concept used. Product rule for differentiating xsin x, then verify the identity using sin²x+cos²x=1.

1. Differentiate y = xsin x using the product rule: y' = ddx(x)sin x + xddx(sin x) = sin x + xcos x.
2. Compute the LHS of the DE: xy' = x(sin x + xcos x) = xsin x + x²cos x.
3. Compute x²-y² using y = xsin x: x²-y² = x²-x²sin²x = x²(1-sin²x) = x²cos²x.
4. Hence √x²-y² = |xcos x|. Under the condition x>y or x<-y (equivalently |y|<|x|, which is true when cos x≠ 0), and taking the principal root in the domain where xcos x≥ 0: √x²-y² = xcos x (up to sign of $xcos x$).
5. Compute the RHS: y + x√x²-y² = xsin x + x· xcos x = xsin x + x²cos x.
6. LHS = xsin x + x²cos x = RHS. Identity verified.

Hence y = xsin x is a solution.

Concept used. Implicit differentiation: differentiate both sides of an implicit equation in x,y with respect to x, treating y as a function of x.

1. Differentiate xy = log y + C implicitly with respect to x. The LHS uses the product rule, the RHS uses the chain rule on log y: ddx(xy) = ddx(log y) + ddx(C). y + x y' = 1y· y' + 0.
2. Collect y' terms on one side: x y' - y'y = -y y'(x - 1y) = -y.
3. Multiply x-1y by yy to combine: y'·xy-1y = -y y' = -y· yxy-1 = -y²xy-1.
4. Multiply numerator and denominator by -1: y' = y²1-xy.
5. This is exactly the given DE (valid for xy≠ 1).

Hence xy = log y + C is a solution.

Concept used. Implicit differentiation and algebraic substitution using the implicit relation itself.

1. Differentiate y - cos y = x implicitly with respect to x: y' - (-sin y) y' = 1 y'(1+sin y) = 1 y' = 11+sin y.
2. From the implicit relation, x = y - cos y, so x + cos y = y.
3. Compute the bracket on the LHS of the DE: ysin y + cos y + x = ysin y + (cos y + x) = ysin y + y = y(1 + sin y).
4. Multiply by y' = 11+sin y (from step 1): (ysin y + cos y + x) y' = y(1+sin y)· 11+sin y = y.
5. LHS = y = RHS. Identity verified.

Hence y - cos y = x is a solution.

Concept used. Implicit differentiation; use ddxtan^-1y = 11+y²· y'.

1. Differentiate x + y = tan^-1y implicitly: 1 + y' = y'1+y².
2. Multiply throughout by (1+y²): (1+y²) + (1+y²) y' = y'.
3. Bring all y' terms to one side: (1+y²) y' - y' = -(1+y²) y' [(1+y²)-1] = -(1+y²).
4. Simplify (1+y²)-1 = y²: y²y' = -(1+y²) y²y' + 1 + y² = 0.
5. This matches the given DE y²y' + y²+1 = 0 exactly.

Hence x + y = tan^-1y is a solution.

Concept used. Differentiate using the chain rule; verify the identity directly.

1. Start with y = (a²-x²)^1/2. Chain rule: y' = 12(a²-x²)^-1/2· ddx(a²-x²) = 12(a²-x²)^-1/2·(-2x).
2. Simplify: y' = -x√a²-x² = -xy.
3. Substitute into the LHS of the DE: x + y y' = x + y·-xy = x - x = 0.
4. LHS = 0 = RHS for every x∈(-a,a) with y≠ 0.

Hence y = √a²-x² is a solution.

Concept used. The general solution of an n-th order differential equation contains exactly n independent arbitrary constants. This is because solving the DE involves n successive integrations, each of which introduces one constant.

1. The order of the given DE is 4.
2. Hence its general solution contains 4 arbitrary constants.

Correct option: (D) 4.

Concept used. A particular solution of a differential equation is obtained from the general solution by assigning particular values to all of its arbitrary constants. By definition, no arbitrary constants remain.

1. A general solution of a third-order DE has 3 arbitrary constants.
2. In a particular solution, every constant has been fixed (typically by initial or boundary conditions).
3. Hence a particular solution contains 0 arbitrary constants.

Correct option: (D) 0.