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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 5 maths textbook solution.

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Answer : \frac{y}{x}=\log\left | x \right |+C

Hint : To solve this equation we use

Give : x\frac{dy}{dx}=x+y

Solution : x\frac{dy}{dx}=1+\frac{y}{x}

              \begin{aligned} &\frac{d y}{d x}-\frac{1}{x} y=1 \\ &\frac{d y}{d x}-\frac{1}{x} y=1 \\ &\frac{d y}{d x}+P y=Q \end{aligned}

              \begin{aligned} &P=\frac{1}{x}, Q=1 \\ &\text { If }=e^{\int P d x} \\ &=e^{-\int \frac{1}{x} d x} \end{aligned}

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

             \begin{aligned} &y \times \frac{1}{x}=\int 1 \times \frac{1}{x} d x+C \\ &\frac{y}{x}=\log x+C \\ &\frac{y}{x}=\log |x|+C \end{aligned}

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