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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 38

Answers (1)

Answer:  y=\frac{x^{2}}{4}+\frac{c}{x^{2}}

Give:  \tan x \frac{d y}{d x}+2 y=x^{2}

Hint: Using  \int \frac{1}{x} d x \text { and } e^{\log e^{x}}=x

Explanation:  \tan x \frac{d y}{d x}+2 y=x^{2}

Divide by x

 \begin{aligned} &=\frac{d y}{d x}+\frac{2 y}{x}=x \\ &=\frac{d y}{d x}=\left(\frac{2}{x}\right) y=x \end{aligned}

This is a linear differential equation of the form

 \begin{aligned} &\frac{d x}{d y}+P y=Q \\ &P=\frac{2}{x} \text { and } Q=x \end{aligned}

The integrating factor If  of this differential equation is

\begin{aligned} &\text { If }=e^{\int \operatorname{Pd} x} \\ &=e^{\int \frac{2}{x} d x} \\ &=e^{2 \int \frac{1}{x} d x} \\ &=e^{2 \log x} \quad\left[\int \frac{1}{x} d x=\log x+C\right] \\ &=e^{\log x^{2}} \\ &=x^{2} \quad\left[e^{\log e^{x}}=x\right] \end{aligned}


Hence, the solution is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y x^{2}=\int x x^{2} d x+C \\ &=y x^{2}=\int x^{3} d x+C \\ &=y x^{2}=\frac{x^{4}}{4}+C \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}


Divide by x^{2}


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