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Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 5

Answers (1)

Answer:  \frac{y}{x}=\log |x|+C

Hint: To solve this equation we use

Give:  x \frac{d y}{d x}=x+y

Solution:  \frac{d y}{d x}=1+\frac{y}{x}

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

\frac{d y}{d x}-\frac{1}{x} y=1 \\

\frac{d y}{d x}+P y=Q

\begin{aligned} &P=\frac{1}{x^{\prime}}\, Q=1 \\ & \end{aligned}

\text { If }=e^{\int P d x} \\

=e^{-\int\frac{1}{x}dx }

=e^{-\log x} \\

=x^{-1} \\


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

\frac{y}{x}=\log x+C \\

\frac{y}{x}=\log |x|+C

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