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 $x = 1$?
Show 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.
Step 1 : Understanding the Question:
The question asks us to find the particular solution to the given first-order linear differential equation under the initial condition $y(1) = 0$. Step 2 : Key Formulas and Approach:
A first-order linear differential equation in standard form is:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
We solve this using the Integrating Factor (I.F.) method:
\[ \text{I.F.} = e^{\int P(x) \, dx} \]
The general solution is then given by:
\[ y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \] Step 3 : Detailed Explanation:
Let us divide the given equation by $x^2$ (since $x > 0$) to bring it into the standard form:
\[ \frac{dy}{dx} + \frac{9}{x}y = x^2 \]
Here, $P(x) = \frac{9}{x}$ and $Q(x) = x^2$.
Let us calculate the Integrating Factor (I.F.):
\[ \text{I.F.} = e^{\int \frac{9}{x} \, dx} = e^{9 \ln x} = e^{\ln x^9} = x^9 \]
Now, write the general solution:
\[ y \cdot x^9 = \int x^2 \cdot x^9 \, dx + C \]
\[ y \cdot x^9 = \int x^{11} \, dx + C \]
\[ y \cdot x^9 = \frac{x^{12}}{12} + C \]
Now, let us apply the initial condition $y = 0$ when $x = 1$:
\[ 0 \cdot (1)^9 = \frac{1^{12}}{12} + C \implies C = -\frac{1}{12} \]
Substitute $C = -\frac{1}{12}$ back into the general solution:
\[ y \cdot x^9 = \frac{x^{12}}{12} - \frac{1}{12} \]
Multiply both sides of the equation by 12:
\[ 12y \cdot x^9 = x^{12} - 1 \]
Now, divide by $x^9$ to solve for $12y$:
\[ 12y = \frac{x^{12} - 1}{x^9} = x^3 - \frac{1}{x^9} \] Step 4 : Final Answer:
The solution of the differential equation is $12y = x^3 - \frac{1}{x^9}$.
This corresponds to Option (A).
