Question

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

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

Hint:

For \(\frac{dy}{dx}+P(x)y=Q(x)\), use integrating factor \(e^{\int P(x)\,dx}\). Here the integrating factor is \(x\).

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{dy}{dx}+\frac{y}{x}=x^2. \] This is a first order linear differential equation of the form: \[ \frac{dy}{dx}+P(x)y=Q(x). \] Here, \[ P(x)=\frac{1}{x}, \qquad Q(x)=x^2. \] The integrating factor is: \[ I.F.=e^{\int P(x)\,dx}. \] So, \[ I.F.=e^{\int \frac{1}{x}\,dx}. \] \[ I.F.=e^{\log x}. \] \[ I.F.=x. \] Now multiply the whole differential equation by \(x\): \[ x\frac{dy}{dx}+y=x^3. \] The left side is the derivative of \(xy\): \[ \frac{d}{dx}(xy)=x^3. \] Now integrate both sides: \[ \int \frac{d}{dx}(xy)\,dx=\int x^3\,dx. \] \[ xy=\frac{x^4}{4}+c. \] Now use the condition: \[ y(1)=1. \] So, when \(x=1\), \[ y=1. \] Substitute: \[ (1)(1)=\frac{1^4}{4}+c. \] \[ 1=\frac{1}{4}+c. \] \[ c=\frac{3}{4}. \] Therefore, \[ xy=\frac{x^4}{4}+\frac{3}{4}. \] Multiply by \(4\): \[ 4xy=x^4+3. \] Hence, the required solution is: \[ 4xy=x^4+3. \]