Question

The solution of the differential equation $x\frac{dy}{dx}+2y=x^2(x\ne 0)$ with $y\left(1\right)=1$, is?

• A. $y=\frac{x^3}{5}+\frac{1}{5x^2}$
• B. $y=\frac{4}{5}{x}^3+\frac{1}{5x^2}$
• C. $y=\frac{3}{4}{x}^2+\frac{1}{4x^2}$
• D. $y=\frac{x^2}{4}+\frac{3}{4x^2}$

Answer 1

Answer: (D)

$y=\frac{x^2}{4}+\frac{3}{4x^2}$

$x\frac{dy}{dx}+2y=x^2:y\left(1\right)=1$
$\frac{dy}{dx}+\left(\frac{2}{x}\right)y=x$ (LDE in y)
IF $=e^{\int \frac{2}{x}dx}=e^{2lnx}=x^2$
$y\cdot (x{)}^2=\int x\cdot x^{2}dx=\frac{x^4}{4}+C$
$y\left(1\right)=1$
$1=\frac{1}{4}+C\Rightarrow C=1-\frac{1}{4}=\frac{3}{4}$
$y{x}^2=\frac{x^4}{4}+\frac{3}{4}$
$y=\frac{x^2}{4}+\frac{3}{4x^2}$.