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  • Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 18 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 18 textbook solution.

Answers (1)

Answer : y \sin x=x^{2} \sin x+C

Hint : To solve this we convert \cot x \text { to } \frac{\cos x}{\sin x} formula.

Give : \frac{d y}{d x}+(\cot x) y=x^{2} \cot x+2 x

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

                \begin{aligned} &P=\cot x, Q=x^{2} \cot x+2 x \\ \end{aligned}

               \begin{aligned} &I f=e^{\int P d x} \\ \end{aligned}

               \begin{aligned} &=e^{\int \cot x d x} \\ \end{aligned}

               \begin{aligned} &=e^{\log \sin x} \\ \end{aligned}

               \begin{aligned} &=\sin x \end{aligned}

               \begin{aligned} &=y I f=\int \text { QIf } d x+C \\ &=y \sin x=\int\left(x^{2} \cot x+2 x\right) \sin x d x+C \\ &=y \sin x=\int x^{2} \frac{\cos x}{\sin x} \sin x+2 x \sin x d x+C \end{aligned}

               =y \sin x=\int x^{2} \cos x d x+\int 2 x \sin x d x+C

               \begin{aligned} &=x^{2} \int \cos x d x-\int \frac{d}{d x} x^{2} \int \cos x d x+2 \int x \sin x d x+C \\ &{\left[\int u v d x=u \int v d x-\int \frac{d y}{d x} \int v d x d x\right]} \end{aligned}

              \begin{aligned} &=x^{2} \sin x-2 \int x \sin x d x+2 \int x \sin x d x+C \\ &=y \sin x=x^{2} \sin x+C \end{aligned}

             

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