Question
Find the general solution of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}+2{\text{y}}=\text{x}^2$
✓
Answer
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}+2{\text{y}}=\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2}{\text{x}}\text{y}=\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\frac{2}{\text{x}}$ and $\text{Q}=\text{x}$
$\therefore \ \text{I}.\text{F}.=\text{e}^{\int{\text{P}{\text{dx}}}}$
$=\text{e}^{\int\frac{2}{\text{x}}{\text{dx}}}$
$={\text{e}^{2\log\text{x}}}$
$=\text{x}^2$
Multiplying both sides of (1) by I.F. = $x^2$, we get
$\text{x}^2\Big(\frac{\text{dy}}{\text{dy}}+\frac{2}{\text{x}}{\text{y}}\Big)={\text{x}^2{\text{x}}}$
$\Rightarrow{\text{x}^2}\frac{{\text{dy}}}{\text{dx}}+2{\text{x}}{\text{y}}={\text{x}^3}$
Integrating both sides with respect to x, we get
$ {\text{x}^2}{\text{y}}=\int{\text{x}^3} \ {\text{dx}}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4}{4}+\text{C}$
$\Rightarrow\text{y}=\frac{\text{x}^2}{4}+\text{C}\text{x}^{-2}$
Hence, $\text{y}=\frac{\text{x}^2}{4}+\text{C}\text{x}^{-2}$ is the required solution.
$\text{x}\frac{\text{dy}}{\text{dx}}+2{\text{y}}=\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{2}{\text{x}}\text{y}=\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\frac{2}{\text{x}}$ and $\text{Q}=\text{x}$
$\therefore \ \text{I}.\text{F}.=\text{e}^{\int{\text{P}{\text{dx}}}}$
$=\text{e}^{\int\frac{2}{\text{x}}{\text{dx}}}$
$={\text{e}^{2\log\text{x}}}$
$=\text{x}^2$
Multiplying both sides of (1) by I.F. = $x^2$, we get
$\text{x}^2\Big(\frac{\text{dy}}{\text{dy}}+\frac{2}{\text{x}}{\text{y}}\Big)={\text{x}^2{\text{x}}}$
$\Rightarrow{\text{x}^2}\frac{{\text{dy}}}{\text{dx}}+2{\text{x}}{\text{y}}={\text{x}^3}$
Integrating both sides with respect to x, we get
$ {\text{x}^2}{\text{y}}=\int{\text{x}^3} \ {\text{dx}}+\text{C}$
$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4}{4}+\text{C}$
$\Rightarrow\text{y}=\frac{\text{x}^2}{4}+\text{C}\text{x}^{-2}$
Hence, $\text{y}=\frac{\text{x}^2}{4}+\text{C}\text{x}^{-2}$ is the required solution.
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