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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 25 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 25 Maths Textbook Solution.

Answers (1)

Answer:\frac{\cos ^{5}}{5}-\frac{\cos ^{3}x}{3}+\left ( x-1 \right )e^{x}+C

Hint:  Apply integration by parts method and formula of\int u\; v\: dx

Given:\frac{dy}{dx}=\sin ^{3}x\cos ^{2}x+xe^{x}

Solution:\frac{dy}{dx}=\sin ^{3}x\cos ^{2}x+xe^{x}

         \Rightarrow \int dy=\int \left ( \sin ^{3}x\cos ^{2}x+xe^{x} \right )dx        (integrate both sides)

        \Rightarrow \int y=\int \sin ^{3}x\cos ^{2}xdx+\int xe^{x}dx

        \Rightarrow y=I_{1}+I_{2}

I_{1}=\int \sin ^{3}x\cos ^{2}xdx

=\int \left ( 1-\cos ^{2}x \right )\cos ^{2}x\sin xdx

Let \cos x=t

        \Rightarrow d \mathrm{t}=-\sin \mathrm{x} \mathrm{d} \mathrm{x} \quad \text { (diff. w.r.t. } \mathrm{x})

I_{1}=-\int t^{2}\left ( 1-t^{2} \right )dt

=\int \mathrm{t}^{2}\left(\mathrm{t}^{2}-1\right) \mathrm{dt}=\int \mathrm{t}^{4}-\mathrm{t}^{2} \mathrm{dt} \\

            \Rightarrow \mathrm{I}_{1}=\frac{\mathrm{t}^{5}}{5}-\frac{\mathrm{t}^{8}}{3}+\mathrm{c}_{1} \\

            \Rightarrow \mathrm{I}_{1}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+\mathrm{c}_{1}

      Now,I_{2}=\int xe^{x}dx

\begin{aligned} &\Rightarrow \mathrm{I}_{2}=x \int \mathrm{e}^{x} \mathrm{dx}-\int \mathrm{e}^{\mathrm{x}} \mathrm{dx} \\ &{\left[\because \int \mathrm{uvd} \mathrm{x}=\mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}\right) \mathrm{dx}\right]} \\ &\Rightarrow \mathrm{I}_{2}=x \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}_{2}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\mathrm{c}_{2} \\ &\text { Now, } \mathrm{y}=\mathrm{I}_{1}+\mathrm{I}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+\mathrm{c}_{1}+(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{c}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{c}\left(\because \mathrm{c}_{1}+\mathrm{c}_{2}=\mathrm{c}\right) \end{aligned}

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