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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 26 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 26 Maths Textbook Solution.

Answers (1)

Answer:\mathrm{c}|2 x-y|^{\frac{5}{8}}\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}=e^{-\frac{3}{8} \tan ^{-1\left(\frac{y}{2 x}\right)}} \\

Given: \left(x^{2}+y^{2}\right) \frac{d y}{d x}=8 x^{2}-3 x y+2 y^{2}

To solve:  We have to solve the given differential equation.

Hint: In homogeneous equation we put x=vy and \frac{dx}{dy}=v+y\frac{dv}{dy}

Solution: Here,

\left(x^{2}+y^{2}\right) \frac{d y}{d x}=8 x^{2}-3 x y+2 y^{2}

\Rightarrow \frac{dx}{dy}=\frac{8x^{2}-3xy+2y^{2}}{\left ( x^{2}+y^{2} \right )}

It is homogeneous equation.

Put y=vy and \frac{dx}{dy}=v+x\frac{dv}{dx}

So, v+x \frac{d v}{d x}=\frac{8 x^{2}-3 x v x+2 x^{2} v^{2}}{x^{2}+x^{2} v^{2}}

\Rightarrow x \frac{d v}{d x}=\frac{8-3 v+2 v^{2}}{1+v^{2}}-v \\

\Rightarrow x \frac{d v}{d x}=\frac{8-3 v+2 v^{2}-v^{3}}{1+v^{2}}-v \\

Separating the variables and integrating we get

\Rightarrow \int \frac{1+v^{2}}{8-4 v+2 v^{2}-v^{3}} d v=\int \frac{1}{x} d x \\

\Rightarrow \int \frac{1+v^{2}}{(2-v)\left(v^{2}+4\right)} d v=\int \frac{1}{x} d x \rightarrow(A) \\

                                                       \Rightarrow \frac{1+v^{2}}{(2-v)\left(v^{2}+4\right)}=\frac{A x+B}{4+v^{2}}+\frac{c}{2-v}[\text { Using partial fraction }] \\

\Rightarrow 1+v^{2}=v^{2}(-A+C)+v(2 A+B)+2 B+4 C

Comparing the coefficient of like power V.

\begin{aligned} &-A+C=1 \rightarrow(1) \\ &2 A-B=0 \\ &\Rightarrow B=2 A \rightarrow(2) \\ &\text { And } 2 B+4 C=1 \rightarrow(3) \end{aligned}

Solving equation \left ( 1 \right ),\left ( 2 \right )& \left ( 3 \right )we get

A=\frac{3}{8},B=\frac{3}{4},C=\frac{5}{8}

Using equation \left ( A \right )we get

\begin{aligned} &\Rightarrow \int \frac{-\frac{3}{8} x-\frac{3}{4}}{4+v^{2}} d v+\frac{5}{8} \int \frac{c}{2-v} d v=\int \frac{d x}{x} \\ &\Rightarrow-\frac{3}{8} \int \frac{v+2}{4+v^{2}} d v+\frac{5}{8} \int \frac{1}{2-v} d v=\int \frac{d x}{x} \end{aligned}

\Rightarrow-\frac{3}{8} \int \frac{v}{4+v^{2}} d v+\frac{2 x 3}{8} \int \frac{1}{4+v^{2}} d v+\frac{5}{8} \int \frac{1}{2-v} d v=\int \frac{d x}{x} \\

\Rightarrow-\frac{3}{16} \log \left|4+v^{2}\right|-\frac{3}{8} \tan ^{-1} \frac{v}{2}+\frac{5}{8} \log |2-v|=\log x+\log c \\

\Rightarrow\left[\log \left[4+v^{2}\right]_{16}^{\frac{8}{16}}+\log e^{\frac{3}{8} \tan ^{-1}\left(\frac{v}{2}\right)}+\log |2-v|^{\frac{5}{8}}\right]=\log c x \\

\Rightarrow\left(4+v^{2}\right)^{\frac{3}{16}} \times e^{\frac{8}{\operatorname{s}} \tan ^{-1}\left(\frac{v}{2}\right)} \times(2-v)^{\frac{5}{8}}=\frac{c}{x}

Put y=vx

\Rightarrow\left(4+\frac{y^{2}}{x^{2}}\right)^{\frac{3}{16}} \times e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \times(2-v)^{\frac{5}{8}}=\frac{c}{x} \\

\Rightarrow\left(\frac{4 x^{2}+y^{2}}{\left(x^{2}\right)^{\frac{8}{16}}}\right)^{\frac{8}{16}} \times \frac{(2-y)^{\frac{5}{8}}}{x^{\frac{5}{8}}}=e^{\frac{a}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \times \frac{c}{x} \\

\Rightarrow \frac{\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}(2-y)^{\frac{5}{8}}}{x}=e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \times \frac{c}{x}\left[\therefore x^{\left.\frac{6}{16}, x^{\frac{5}{8}}=x^{\frac{6}{16}}+\frac{5}{8}=x^{\frac{8}{8}}=x\right]}\right. \\

\Rightarrow\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}(2-y)^{\frac{5}{8}}=c e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \\

\Rightarrow \mathrm{c}|2 x-y|^{\frac{5}{8}}\left(4 x^{2}+y^{2}\right)^{\frac{3}{16}}=e^{\frac{3}{8} \tan ^{-1}\left(\frac{y}{2 x}\right)} \text { where } c=\frac{1}{c}

This is required solution.

Posted by

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