Vikolers6

2021-12-19

Marcus Herman

Expert

Step 1
we have to solve the given differential equations with intial condition.
We can solve it variable seperable
Step 2
Given
$⇒2ydx=x{y}^{3}dy-3xdy$
$⇒2ydx=x\left({y}^{3}-3\right)dy$
$⇒\int \frac{2dx}{x}=\int \left(\frac{{y}^{3}-3}{y}\right)dy$
$⇒2\int \frac{dx}{x}=\int \left({y}^{2}-\frac{3}{y}\right)dy$
$⇒2\mathrm{ln}\left(x\right)=\frac{{y}^{3}}{3}-3\mathrm{ln}\left(y\right)+c$ (1)
Put $x=1,y=1$
$⇒2\mathrm{ln}\left(1\right)=\frac{{1}^{3}}{3}-3\mathrm{ln}\left(1\right)+c$
$⇒0=1-0+c$
$\therefore c=-1$
Put
$2\mathrm{ln}\left(x\right)=\frac{{y}^{3}}{3}-3\mathrm{ln}\left(y\right)-1$

amarantha41

Expert

We need to solve $2ydx+3xdy=x{y}^{3}dy,y\left(1\right)=1$
So, $2ydx+3xdy=x{y}^{3}dy$
$⇒2y+3x\frac{dy}{dx}=x{y}^{3}\frac{dy}{dx}$
Substitute $\frac{dy}{dx}$ with y

RizerMix

Expert

$⇒2ydx=\left(x{y}^{3}-3x\right)dy$
$⇒2ydx=x\left({y}^{3}-3\right)dy$
Now using sepration of variable method then
$⇒2\frac{dx}{x}=\frac{\left({y}^{3}-3\right)}{y}dy$
Integrating both side then
$2\int \frac{dx}{x}=\int \left({y}^{2}-\frac{3}{y}\right)dy$
$2\mathrm{ln}x=\frac{{y}^{3}}{3}-3\mathrm{ln}y+c$ (1)
Now, given initial condition x=1, y=1
Now by Eqn (1)
$2\mathrm{ln}\left(1\right)=\frac{1}{3}-3\mathrm{ln}1+c$
$2×0=\frac{1}{3}-0+c$
$c=\frac{-1}{3}$
Put value of c in eq-n (1) then
$2\mathrm{ln}x=\frac{{y}^{3}}{3}-3\mathrm{ln}y-\frac{1}{3}$
$\frac{{y}^{3}}{3}-3\mathrm{ln}y=2\mathrm{ln}x+\frac{1}{3}$

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