differential equation

Question

(x^3+3xy^2)dy/dx=y^3+3x^2y

Answer 1

\((x^3+3xy^2)\frac{\mathrm dy}{\mathrm d x} = y^3+3x^2y\)
\(\frac{\mathrm dy}{\mathrm d x} =\frac{y^3+3x^2y}{x^3+3xy^2}\)

\(F(x,y)=\frac{y^3+3x^2y}{x^3+3xy^2}\)

Let \(x=\lambda x\) and \(y=\lambda y\) and replace by \(x=\lambda x\) and \(y=\lambda y\)

\(F(\lambda x,\lambda y)=\frac{(\lambda y)^3+3(\lambda x)^2 \lambda y}{(\lambda x)^3+3\lambda x(\lambda y)^2}\)

\(F(\lambda x,\lambda y)=\frac{\lambda^3 y^3+3\lambda^2 x^2 \lambda y}{\lambda^3 x^3+3\lambda x\lambda^2 y^2}\)

\(F(\lambda x,\lambda y)=\frac{\lambda^3 y^3+3\lambda^3 x^2 y}{\lambda^3 x^3+3\lambda^3 x y^2}\)

\(F(\lambda x,\lambda y)=\frac{\lambda^3( y^3+3 x^2 y)}{\lambda^3 (x^3+3x y^2)}\)

\(F(\lambda x,\lambda y)=\frac{( y^3+3 x^2 y)}{(x^3+3x y^2)}\)

\(F(\lambda x,\lambda y)=\lambda^0 \frac{( y^3+3 x^2 y)}{(x^3+3x y^2)}\)

This is a  zero degree homogeneous differential equation. 

Let \(y=vx\)

Differentiating with respect to x, we will get ,

\(\frac{\mathrm dy}{\mathrm d x} =v+x\frac{\mathrm dv}{\mathrm d x} \)

\(v+x\frac{\mathrm dv}{\mathrm d x} =\frac{(vx)^3+3x^2vx}{x^3+3x(vx)^2}\)

\(v+x\frac{\mathrm dv}{\mathrm d x} =\frac{v^3x^3+3vx^3}{x^3+3vx^3}\)

\(v+x\frac{\mathrm dv}{\mathrm d x} =\frac{x^3(v^3+3v)}{x^3(1+3v)}\)

\(x\frac{\mathrm dv}{\mathrm d x} =\frac{(v^3+3v)}{(1+3v)}-v\)

\(x\frac{\mathrm dv}{\mathrm d x} =\frac{(v^3+3v)-v(1+3v)}{(1+3v)}\)

\(x\frac{\mathrm dv}{\mathrm d x} =\frac{v^3+3v-v-3v^2}{(1+3v)}\)

\(x\frac{\mathrm dv}{\mathrm d x} =\frac{v^3+2v-3v^2}{(1+3v)}\)

\(x\frac{\mathrm dv}{\mathrm d x} =\frac{v(v^2-3v+2)}{(1+3v)}\)

\(\frac{(1+3v)}{v(v^2-3v+2)} dv=\frac{dx}{x}\)

\(\frac{(1+3v)}{v(v-2)(v-1)} dv=\frac{dx}{x}\)

Using partial fraction, 

\((\frac{1}{2v}+\frac{7}{2(v-2)}-\frac{4}{(v-1)} )dv=\frac{dx}{x}\)

\(\int (\frac{1}{2v}+\frac{7}{2(v-2)}-\frac{4}{(v-1)} )dv=\int \frac{dx}{x}\)

\(\int \frac{1}{2v}dv+\int \frac{7}{2(v-2)}dv-\int \frac{4}{(v-1)} dv=\int \frac{dx}{x}\)

\( \frac{1}{2}ln|v|+ \frac{7}{2}ln|v-2|- \frac{1}{4}ln|v-1| =ln|x|+c\)

\(\frac{1}{2}ln|\frac{y}{x}|+ \frac{7}{2}ln|\frac{y}{x}-2|- \frac{1}{4}ln|\frac{y}{x}-1| =ln|x|+c\)

Answer 2

\(\frac {dy}{dx} = \frac{^3 + 3x^2y}{x^3 + 3xy^2}\)

Let \(y = vx\)

\(\therefore \frac{dy}{dx} = x \frac{dv}{dx} + v\)

\(v + x \frac {dv}{dx} = \frac{v^3 + 3v}{1 +3v^2}\)

\(\therefore x \frac {dv}{dx} = \frac {v^3 + 3v - v - 3v^3}{1 + 3v^2}\)

⇒ \(\frac {1 + 3v^2}{2v - 2v^3} dv = \frac {dx}x\)

⇒ \(\frac 12 \int \frac{(1 + 3v^2)dv}{v - 3} = \int \frac {dx}x\)

⇒ \(\frac 12 \int \frac {1 - 3v^2 + 6v^2}{v - v^3}dv = \log x + \log c\)

⇒ \(\frac 12 \log (v - v^3) + 3 \int \frac{v^2}{v(1 -v)(1 +v) }dv = \log (cx)\)

⇒ \(\log (v - v^3) - 6 \int \frac {v dv}{(v -1)(v + 1)} = \log(c^2x^2)\)

⇒ \(\log (v - v^3) -\frac 62 \int \left(\frac 1{v-1} + \frac 1{v + 1}\right) dv = \log (c^2 x^2)\)

⇒ \(\log (v - v^3) - 3\log ((v-1) (v + 1)) = \log (c^2 x^2)\)

⇒ \(\log \frac{v -v^3}{(v - 1)^3 (v + 1)^3} = \log (c^2x^2)\)

 ⇒ \(\frac{x^2 y - y^3}{(y - x)^3 (y +x)^3} = c^2x^5\)

⇒ \(y (x^2 - y^2) = c^2 x^5 (x + y)^3 (y -x)^3\)

⇒ \(y (x - y) (x + y) = - c^2 x^5 (x + y)^3 (x - y)^3\)

⇒ \(y = -c^2x^5 (x +y)^2 (x - y)^2\)