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

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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}


              \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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