Question

guys can anyone help me to do this in homogeneous

Answer 1

yea sup

Answer 2

y(8x-9y)dx+2x(x-3y)dy=0

Answer 3

or by I.F. will do ^.^

Answer 4

idont know sorry

Answer 5

@DanJS @Nnesha

Answer 6

Given the ODE
\[y(8x-9y)\,\mathrm dx+2x(x-3y)\,\mathrm dy=0\]you can rewrite it as
\[\frac{\mathrm dy}{\mathrm dx}=\frac{y(9y-8x)}{2x(x-3y)}=f(x,y)\]The ODE is homogeneous if for some integer \(\alpha\) and nonzero \(t\), you have \(f(tx,ty)=t^\alpha f(x,y)\). Check:
\[\begin{align*}
f(tx,ty)&=\frac{ty(9ty-8x)}{2tx(tx-3ty)}\\[1ex]
&=\frac{t^2y(9y-8x)}{2t^2x(x-3y)}\\[1ex]
&=\frac{y(9y-8x)}{2x(x-3y)}\\[1ex]
&=t^0f(x,y)
\end{align*}\]Okay, now that you know the ODE is homogeneous, you can try applying the usual substitution \(y=vx\), which makes \(\mathrm dy=v\,\mathrm dx+x\,\mathrm dv\), so the original ODE can be rewritten as
\[\begin{align*}
vx(8x-9vx)\,\mathrm dx+2x(x-3vx)(v\,\mathrm dx+x\,\mathrm dv)&=0\\[1ex]
(10x^2v-15v^2x^2)\,\mathrm dx+(2x^3-6vx^3)\,\mathrm dv&=0\\[1ex]
(10v-15v^2)\,\mathrm dx+(2x-6vx)\,\mathrm dv&=0\\[1ex]
\frac{\mathrm dv}{\mathrm dx}&=\frac{5v(3v-2)}{2x-6vx}\\[1ex]
\frac{\mathrm dv}{\mathrm dx}&=\frac{15v^2-10v}{2x(1-3v)}
\end{align*}\]which is separable.

Answer 7

Those last two steps should have been
\[\frac{\mathrm dv}{\mathrm dx}=\frac{15v^2-10v}{2x-6vx}\implies\frac{\mathrm dv}{\mathrm dx}=\frac{5v(3v-2)}{2x(1-3x)}\]Sorry for any confusion!

Answer 8

If you want the "textbook" way of solving this equation ...

\(\color{black}{\displaystyle y(8x-9y)dx+2x(x-3y)dy=0 }\)

\(\color{black}{\displaystyle M(x,y):=8xy-9y^2 }\)
\(\color{black}{\displaystyle N(x,y):=2x^2-6xy }\)

\(\color{black}{\displaystyle \frac{dM}{dy}=8x-18y }\) \(\\[1.4em]\)
\(\color{black}{\displaystyle \frac{dN}{dx}=4x-6y }\)

So, the equation is not exact.
Try INTEGRATING FACTOR.

\(\color{black}{\displaystyle \xi(x)=\frac{M_y-N_x}{N} }\) \(\\[1.4em]\)
\(\color{black}{\displaystyle \xi(x)=\frac{(8x-18y)-(4x-6y)}{2x^2-6xy} }\) \(\\[1.4em]\)
\(\color{black}{\displaystyle \xi(x)=\frac{2(2x-6y)}{x(2x-6y)}=\frac{2}{x} }\)

\(\color{black}{\displaystyle \mu(x)=\exp\left\{ \int\xi(x)~dx \right\}=\exp\left\{ \int \frac{2}{x} ~dx \right\}=e^{2\ln(x)} =x^2 }\)
((Remark: I omit the integration constant, b/c we are looking for an integrating factor, not a family of functions with *this* derivative))

\(\color{black}{\displaystyle (\color{red}{x^2})y(8x-9y)dx+(\color{red}{x^2})2x(x-3y)dy=0 }\)
\(\color{blue}{\displaystyle (8yx^3-9y^2x^2)dx+(2x^4-6x^3y)dy=0 }\) \(\color{red }{\bf \Leftarrow\Leftarrow~Your~~NEW~~Equation }\)

\(\color{black}{\displaystyle M_2(x,y):=8yx^3-9y^2x^2 }\)
\(\color{black}{\displaystyle N_2(x,y):=2x^4-6x^3y }\)
(subscripts are there to denote the new functions M and N)

\(\color{black}{\displaystyle \frac{dM}{dy}=8x^3-18yx^2 }\) \(\\[1.4em]\)
\(\color{black}{\displaystyle \frac{dN}{dx}=8x^3-18x^2y }\)

Answer 9

then, proceed the regular way ...