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Please solve RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 3 maths textbook solution.

 

 

 

 

 

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Answer: y+x^2=\frac{e^{3x}}{3}+C

Hint: You have to integrate by applying integration of xn.

Given: \frac{dy}{dx}+2x=e^{3x}

Solution:\frac{dy}{dx}+2x=e^{3x}

\frac{dy}{dx}=-2x +e^{3x}

dy=\left (-2x +e^{3x} \right )dx

Integrating both sides,

\begin{aligned} &\int d y=\int\left(-2 x+e^{3 x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \frac{1}{x} d x=\log |x|+C\right] \\ &y=-\frac{2 x^{2}}{2}+\frac{e^{a x}}{3} \\ &\quad=-x^{2}+\frac{e^{3 x}}{3} \\ &y+x^{2}=\frac{e^{3 x}}{3}+C \end{aligned}

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