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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 24 Maths Textbook Solution.

Answers (1)

Answer: $x=2 y^{3}+C y^{-2}$

Hint: To solve this equation we use formula.

Give: $\begin{aligned} &\left(2 x-10 y^{3}\right) \frac{d y}{d x}+y=0 \\ & \end{aligned}$

Solution: $\frac{d y}{d x}=-\frac{y}{2 x-10 y^{3}}$

$\begin{aligned} &=\frac{d y}{d x}=\frac{-\left(2 x-10 y^{3}\right)}{y} \\ & \end{aligned}$

$=\frac{d y}{d x}=-\frac{2 x}{y}+\frac{10 y^{3}}{y} \\$

$=\frac{d x}{d y}+\frac{2 x}{y}=10 y^{2} \\$

$=\frac{d x}{d y}+R x=S \\$

$R=\frac{2 x}{y}, S=10 y^{2}$

$\begin{aligned} &I f=e^{\int R d y} \\ & \end{aligned}$

$=e^{2 \int \frac{1}{y} d y}$

$\begin{aligned} &=e^{2 \log y} \\ & \end{aligned}$

$=y^{2} \\$

$=x I f=\int S I f+C \\$

$=x y^{2}=\int S I f+C \\$

$=x y^{2}=\int 10 y^{2} y^{2}+C$

$\begin{aligned} &=x y^{2}=\int 10 y^{4}+C \\ &=x y^{2}=\frac{10 y^{5}}{5}+C \\ & \end{aligned}$

$=x y^{2}=2 y^{5}+C \\$

$=x=2 y^{3}+\frac{C}{y^{2}} \\$

$=x=2 y^{3}+C y^{-2}$

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