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In each of the following differential equation show that it is homogeneous and solve it. \left(x^{3}+3 x y^{2}\right) d x+\left(y^{3}+3 x^{2} y\right) d y=0
CBSE | Class 12 | exercise 20 | Homogeneous Differential Equations | Maths | RS Aggarwal
In each of the following differential equation show that it is homogeneous and solve it. \left(x^{3}+3 x y^{2}\right) d x+\left(y^{3}+3 x^{2} y\right) d y=0

Solution:

\begin{array}{l} \Rightarrow\left(x^{3}+3 x y^{2}\right) d x+\left(y^{3}+3 x^{2} y\right) d y=0 \\ \left.\left.\Rightarrow \frac{d y}{d x}=-\frac{\left(x^{3}+3 x y^{2}\right)}{\left(y^{2}+3 x^{2} y\right)}=-\frac{3 x y^{2}\left(\frac{x^{3}}{3 x y^{2}}+1\right)}{3 x^{2} y} \frac{y^{3}}{3 x^{2} y}+1\right)=-\frac{y\left(\frac{x^{2}}{3 y^{2}}+1\right)}{x} \frac{y^{2}}{3 x^{2}}+1\right) \\ \Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right) \end{array}
\Rightarrow the given differential eq. is a homogenous eq.
Solution of the given differential eq. is:
\begin{array}{l} \text { Putting } y=v x \\ \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \begin{aligned} v+x \frac{d v}{d x}=\left.{\left.-\frac{v x}{x} \frac{\left(v^{2}\right.}{3(v x)^{2}}+1\right)}{\left(\frac{(v x)^{2}}{3 x^{2}}+1\right)}=-V \frac{(v)}{(v)^{2}}{(v)^{2}}+1\right) \end{aligned} \\ =-\frac{1+3(v)^{2}}{3 v+(v)^{3}} \\ \Rightarrow x \frac{d v}{d x}=-\frac{1+3(v)^{2}}{3 v+(v)^{3}}-v=-\frac{1+3(v)^{2}+3(v)^{2}+(v)^{4}}{3 v+(v)^{3}} \\ =\frac{1+6(v)^{2}+(v)^{4}}{3 v+(v)^{3}} \\ \Rightarrow \frac{3 v+(v)^{3}}{1+6(v)^{2}+(v)^{4}} d v=-\frac{d x}{x} \end{array}
On integrating both sides we obtain:
\begin{array}{l} \Rightarrow \int \frac{3 v+(v)^{3}}{1+6(v)^{2}+(v)^{4}} d v=-\int \frac{d x}{x}+c \\ \Rightarrow \frac{\ln \left|1+6(v)^{2}+(v)^{4}\right|}{4}+\ln |x|=\ln |c| \end{array}
By resubstituting the value of \mathrm{y}=\mathrm{v} \mathrm{x} we have
\begin{array}{l} \Rightarrow \frac{\ln \left|1+6\left(\frac{y}{x}\right)^{2}+\left(\frac{y}{x}\right)^{4}\right|}{4}+\ln |x|=\ln |c| \\ \Rightarrow y^{4}+6 x^{2} y^{2}+x^{4}=C \end{array}

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