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

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}

=y=\frac{x^{2}}{4}+\frac{c}{x^{2}} 

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