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  • Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xiv) textbook solution.

Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xiv) textbook solution.

Answers (1)

Answer : x=\frac{y^{3}}{3}+\frac{c}{y}

Hint : integrate by applying integration of x^{n}

Given : y \; d x+\left(x-y^{2}\right) d y=0

Solution : y \; d x+\left(x-y^{2}\right) d y=0

                \frac{d y}{d x}=\frac{-y}{x-y^{2}}

This is not in the form of \frac{d y}{d x}+Py=Q

\begin{aligned} &\frac{d x}{d y}=\frac{y^{2}-x}{y} \\ \end{aligned}

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

\begin{aligned} &\frac{d x}{d y}+\frac{x}{y}=y \end{aligned}                                               .....(i)

Find P and Q Where P=\frac{1}{y} \quad, Q=y

Find I.F

\begin{aligned} \text { I.F } &=e^{\int P d y} \\ &=e^{\int \frac{1}{y} d y}=e^{\log y}=y \end{aligned}

Solution will be

\begin{aligned} &x(I . F)=\int(Q \times I . F) d y+c \\ &x y=\int y \times y d y+c \\ &x y=\int y^{2} d y+c \end{aligned}

\begin{aligned} x y &=\frac{y^{3}}{3}+c \\ x &=\frac{y^{2}}{3}+\frac{c}{y} \end{aligned}

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