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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 10 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 10 textbook solution.

Answers (1)

Answer : y=e^{x}+Cx

Hint: To solve this equation we will use differentiate separately.

Give : \begin{aligned} &x \frac{d y}{d x}-y=(x-1) e^{x} \\ \end{aligned}

Solution : \begin{aligned} & \frac{d y}{d x}-\frac{1}{x} y \frac{(x-1) e^{x}}{x} \end{aligned}

              \frac{dy}{dx}+Py=Q

              \begin{aligned} &P=\frac{-1}{x}, Q=\frac{(x-1) e^{x}}{x} \\ &I f=e^{\int P d x} \\ &=e^{-\int \frac{1}{x} d x} \end{aligned}

             \begin{aligned} &=e^{\log x^{-1}} \\ &=x^{-1} \end{aligned}

            \begin{aligned} &y I f=\int Q I f+C \\ &=y \frac{1}{x}=\int(x-1) e^{x} \frac{1}{x} d x+C \\ &=\frac{y}{x}=\int \frac{x e^{x}}{x^{2}} d x-\int \frac{e^{x}}{x^{2}} d x+C \end{aligned}

             \begin{aligned} &=\frac{1}{x} e^{x}-\int \frac{1}{x^{2}} e^{x} d x-\int \frac{1}{x^{2}} e^{x} d x \\ &=\frac{1}{x} e^{x} \\ &=\frac{y}{x}=\frac{e^{x}}{x}+C \\ &=y=e^{x}+C x \end{aligned}

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