Question

Which one of the following is the solution of the differential equation
\[ x^2 \frac{dy}{dx} + 9xy = x^4 \quad (\text{for } x > 0) \]
given that $y = 0$ when $x = 1$?

• A. $12y = x^3 - \frac{1}{x^9}$
• B. $12y = x^9 - \frac{1}{x^3}$
• C. $9y = x^{21} - \frac{1}{x^3}$
• D. $9y = x^3 - \frac{1}{x^{21}}$

Hint:

Before doing the integration, always divide out the coefficient of $\frac{dy}{dx}$ to ensure the equation is in standard form.
Failing to do so will result in an incorrect Integrating Factor.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Divide by $x^2$ to get the standard linear form: $\frac{dy}{dx} + \frac{9}{x}y = x^2$. This is $\frac{dy}{dx} + Py = Q$.
Step 2: Finding the Integrating Factor (I.F.):
$I.F. = e^{\int \frac{9}{x} dx} = e^{9 \ln x} = x^9$.
Step 3: Solving the Equation:
$y \cdot (I.F.) = \int Q \cdot (I.F.) dx$ $y \cdot x^9 = \int x^2 \cdot x^9 dx = \int x^{11} dx$ $y x^9 = \frac{x^{12}}{12} + C$.
Step 4: Applying Initial Conditions:
At $x=1, y=0$: $0 = 1/12 + C \implies C = -1/12$. $y x^9 = \frac{x^{12} - 1}{12} \implies 12y = \frac{x^{12} - 1}{x^9} = x^3 - \frac{1}{x^9}$.
Final Answer:
The solution is 12y = x\(^3\) - 1/x\(^9\).