The solution of the differential equation

Question:

The solution of the differential equation

$x \frac{d y}{d x}+2 y=x^{2}(x \neq 0)$ with $y(1)=1$, is

 

  1. $y=\frac{x^{3}}{5}+\frac{1}{5 x^{2}}$

  2. $y=\frac{4}{5} x^{3}+\frac{1}{5 x^{2}}$

  3. $y=\frac{3}{4} x^{2}+\frac{1}{4 x^{2}}$

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


Correct Option: , 4

Solution:

$x \frac{d y}{d x}+2 y=x^{2}: y(1)=1$

$\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{2}{\mathrm{x}}\right) \mathrm{y}=\mathrm{x}$ (LDE in $\left.\mathrm{y}\right)$

$I F=e^{\int \frac{2}{x} d x}=e^{2(n x}=x^{2}$

$y \cdot\left(x^{2}\right)=\int x \cdot x^{2} d x=\frac{x^{4}}{4}+C$

$\mathrm{y}(1)=1$

$1=\frac{1}{4}+\mathrm{C} \Rightarrow \mathrm{C}=1-\frac{1}{4}=\frac{3}{4}$

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

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

 

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