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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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