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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 11

Answers (1)

Answer: $5 x y=e^{5}+C$

Hint: To solve this equation we use $e^{\int P dx}$ formula.

Give: $\begin{aligned} &x \frac{d y}{d x}+\frac{y}{x}=x^{3} \\ & \end{aligned}$

Solution: $\frac{d y}{d x}+P y=Q$

$\begin{aligned} &P=\frac{1}{x^{\prime}} Q=x^{3} \\ & \end{aligned}$

$I\! f=e^{\int P d x} \\$

$=e^{\int \frac{1}{x} d x} \\$

$=e^{\log x} \\$

$=x \\$

$y I \! f=\int Q I\! f+C$

$\begin{aligned} &=y x=\int x^{3} x d x+C \\ & \end{aligned}$

$=y x=\int x^{4} d x+C \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\$

$=x y=\frac{x^{5}}{5}+C \\$

$=y=\frac{x^{4}}{5}+\frac{C}{x}$

$=5 x y=x^{5}+C$

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