Question

The solution of the differential equation \[ \frac{dy}{dx}+\frac{y}{x}=x^2 \] under the condition \(y(1)=1\) is

• A. \(4xy=x^3+3\)
• B. \(4xy=x^4+3\)
• C. \(4xy=x^3-3\)
• D. \(4xy=x^4-3\)

Hint:

For linear differential equation, use integrating factor \(e^{\int P(x)\,dx}\).

Answer 1

The Correct Option is B

Step 1: Given: \[ \frac{dy}{dx}+\frac{1}{x}y=x^2 \] This is linear differential equation.

Step 2: Integrating factor: \[ I.F.=e^{\int \frac{1}{x}\,dx}=e^{\log x}=x \]

Step 3: Multiply by \(x\): \[ x\frac{dy}{dx}+y=x^3 \] \[ \frac{d}{dx}(xy)=x^3 \]

Step 4: Integrate: \[ xy=\frac{x^4}{4}+c \] \[ 4xy=x^4+4c \]

Step 5: Use \(y(1)=1\): \[ 4(1)(1)=1+4c \] \[ 4c=3 \]

Step 6: \[ 4xy=x^4+3 \] \[ \boxed{4xy=x^4+3} \]