Solve the following differential equation: dy dx + y x =x^3 - Vidyadip

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^3$

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Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{x}^3\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=-\frac{1}{\text{x}}$
$\text{Q}=\text{x}^3$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{\log|\text{x}|}=\text{x}$
Multiplying both sides of (1) by x, we get
$\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}\Big)=\text{x x}^3$
$\Rightarrow\ \text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}^4$
Integrating both sides with respect to x, we get
$\text{xy}=\int\text{x}^4\text{dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^5}{5}+\text{C}$
$\Rightarrow\ 5\text{xy}=\text{x}^5+5\text{C}$
$\Rightarrow\ 5\text{xy}=\text{x}^5+\text{K}$ (where, K = 5C)
Hence, $5\text{xy}=\text{x}^5+\text{K}$ is the required solution.

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