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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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