Q.For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

Concept: Verification of Solution — substitute the given function into the differential equation and check that it satisfies the identity.

(i) $xy = a e^x + b e^{-x} + x^2$

Differentiate twice:

$y + x y' = a e^x - b e^{-x} + 2x$

$2y' + x y'' = a e^x + b e^{-x} + 2$

Now $x y'' + 2 y' - x y + x^2 - 2 = (a e^x + b e^{-x} + 2) - x y + x^2 - 2 = (a e^x + b e^{-x} + x^2) - x y = 0$ since $x y = a e^x + b e^{-x} + x^2$. Verified.

(ii) $y = e^x (a \cos x + b \sin x)$

$y' = e^x (a \cos x + b \sin x) + e^x (-a \sin x + b \cos x) = y + e^x (-a \sin x + b \cos x)$

$y'' = y' + e^x (-a \sin x + b \cos x) + e^x (-a \cos x - b \sin x) = y' + (y' - y) - y = 2y' - 2y$

Thus $y'' - 2y' + 2y = 0$. Verified.

(iii) $y = x \sin 3x$

$y' = \sin 3x + 3x \cos 3x$

$y'' = 3 \cos 3x + 3 \cos 3x - 9x \sin 3x = 6 \cos 3x - 9y$

Then $y'' + 9y - 6 \cos 3x = (6 \cos 3x - 9y) + 9y - 6 \cos 3x = 0$. Verified.

(iv) $x^2 = 2y^2 \log y$

Differentiate implicitly: $2x = 4y \log y \cdot y' + 2y^2 \cdot \frac{1}{y} y' = 4y y' \log y + 2y y'$

So $2x = 2y y' (2 \log y + 1)$ → $y' = \frac{x}{y (2 \log y + 1)}$

Now $(x^2 + y^2) y' - x y = \left(2y^2 \log y + y^2\right) \cdot \frac{x}{y (2 \log y + 1)} - x y = \frac{y^2 (2 \log y + 1) \cdot x}{y (2 \log y + 1)} - x y = x y - x y = 0$. Verified.

(i) $xy = a e^x + b e^{-x} + x^2$ : $x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$

Concept: The given relation is implicit in $y$. We can either solve for $y$ explicitly or differentiate the equation as it stands. Since $y$ appears multiplied by $x$, solving explicitly is straightforward: $y = \frac{a e^x + b e^{-x} + x^2}{x}$. Then we compute $y'$ and $y''$ and substitute.

1. Write $y$ explicitly:

2. First derivative (using quotient rule on the first two terms, or rewrite as $a e^x x^{-1}$ and differentiate):

Factor common terms:

3. Second derivative: Differentiate $y'$ term by term. For the first term, use quotient rule on $\frac{a e^x (x-1)}{x^2}$:

   • Let $u = a e^x (x-1)$, $v = x^2$. Then $u' = a e^x (x-1) + a e^x = a e^x x$, and $v' = 2x$.
   • So derivative = $\frac{(a e^x x)(x^2) - (a e^x (x-1))(2x)}{x^4} = \frac{a e^x x^3 - 2a e^x x(x-1)}{x^4} = \frac{a e^x (x^2 - 2x + 2)}{x^3}$

   For the second term, $-\frac{b e^{-x} (x+1)}{x^2}$:

   • $u = -b e^{-x}(x+1)$, $v = x^2$. $u' = -b[-e^{-x}(x+1) + e^{-x}] = -b e^{-x}(-x) = b e^{-x} x$ (careful: derivative of $e^{-x}(x+1)$ is $-e^{-x}(x+1) + e^{-x} = -x e^{-x}$, so $u' = -b(-x e^{-x}) = b x e^{-x}$).
   • Derivative = $\frac{(b x e^{-x})(x^2) - (-b e^{-x}(x+1))(2x)}{x^4} = \frac{b e^{-x} x^3 + 2b e^{-x} x(x+1)}{x^4} = \frac{b e^{-x} (x^2 + 2x + 2)}{x^3}$

   The derivative of $1$ is $0$. So:

4. Substitute into the differential equation: $x y'' + 2 y' - x y + x^2 - 2 = 0$

   Compute $x y''$:

Compute $2 y'$:

Compute $-x y$:

Now sum everything:

Combine the $a e^x$ terms: factor $\frac{a e^x}{x^2}$:

Combine the $b e^{-x}$ terms: factor $\frac{b e^{-x}}{x^2}$:

The constant terms: $+2 - x^2 + x^2 - 2 = 0$.

Everything cancels. Hence the given function satisfies the differential equation.

(ii) $y = e^x (a \cos x + b \sin x)$ : $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$

Concept: This is a linear combination of $e^x \cos x$ and $e^x \sin x$, which are known to satisfy the second-order linear ODE $y'' - 2y' + 2y = 0$. We verify by direct differentiation.

1. First derivative: Use product rule. Let $u = e^x$, $v = a \cos x + b \sin x$.

2. Second derivative: Differentiate $y'$ again. Write $y' = e^x[(a+b)\cos x + (b-a)\sin x]$. Apply product rule:

Simplify:

3. Substitute into $y'' - 2y' + 2y$:

Factor $e^x$ and collect $\cos x$ and $\sin x$ terms:

For $\cos x$: $2b - 2(a+b) + 2a = 2b - 2a - 2b + 2a = 0$

For $\sin x$: $-2a - 2(b-a) + 2b = -2a - 2b + 2a + 2b = 0$

Hence the expression is identically zero.

(iii) $y = x \sin 3x$ : $\frac{d^2 y}{dx^2} + 9y - 6 \cos 3x = 0$

Concept: Here $y$ is a product of $x$ and $\sin 3x$. We compute two derivatives and substitute. The presence of $-6\cos 3x$ in the equation suggests that after substitution, the $\sin$ terms will cancel and leave a $\cos$ term that matches.

1. First derivative: Using product rule:

2. Second derivative: Differentiate $y'$:

3. Substitute into $y'' + 9y - 6 \cos 3x$:

The $6 \cos 3x$ and $-6 \cos 3x$ cancel. The $-9x \sin 3x + 9x \sin 3x$ also cancel. Result is $0$.

(iv) $x^2 = 2y^2 \log y$ : $(x^2 + y^2) \frac{dy}{dx} - xy = 0$

Concept: This relation is implicit. We differentiate both sides with respect to $x$, treating $y$ as a function of $x$, then solve for $\frac{dy}{dx}$ and substitute into the given equation.

1. Differentiate the given relation implicitly:

Left side: $2x$.

Right side: $2 \cdot \frac{d}{dx}(y^2 \log y)$. Use product rule: derivative of $y^2$ is $2y \frac{dy}{dx}$, derivative of $\log y$ is $\frac{1}{y} \frac{dy}{dx}$.

So:

Hence:

2. Solve for $\frac{dy}{dx}$:

3. Substitute into $(x^2 + y^2) \frac{dy}{dx} - xy$:

Factor $x$:

Combine inside the bracket over a common denominator:

4. Use the original relation: $x^2 = 2y^2 \log y$. Substitute $x^2$:

Hence the numerator is zero, so the entire expression is zero.

Method: Verifying a given function solves a differential equation

To verify (rather than solve), differentiate the given function enough times, substitute, and show the equation reduces to an identity.

Steps

Step 1: Count how many derivatives you need.

Look at the highest derivative in the equation; you must compute up to that order.

Step 2: Differentiate carefully.

Use product, chain and — for an implicit relation like $x^2=2y^2\log y$ — implicit differentiation, keeping $\dfrac{dy}{dx}$ symbolic.

Step 3: Substitute into the equation.

Replace each derivative and $y$ by your expressions.

Step 4: Simplify to $0=0$.

Use the original relation to eliminate leftover terms; reaching an identity confirms it is a solution. You need not derive the solution from scratch.

Common Mistakes

Mistake 1: Trying to solve the equation instead of verifying.

Why it's wrong: the task is to substitute the given function and confirm an identity, not to derive it. Correct approach: differentiate the given $y$ up to the required order and plug in.

Mistake 2: Chain-rule and product-rule slips.

Why it's wrong: e.g. for $y=x\sin 3x$, $y''=6\cos 3x-9x\sin 3x$ (the factor $3$ from $\sin 3x$ is essential); missing it breaks the cancellation. Correct approach: differentiate carefully term by term.

Mistake 3: Not using the original relation for an implicit case.

Why it's wrong: in (iv) the numerator only vanishes after substituting $x^2=2y^2\log y$. Correct approach: use the given relation to eliminate leftover terms and reach $0=0$.