Vikolers6

Answered

2021-12-19

Answer & Explanation

Marcus Herman

Expert

2021-12-20Added 41 answers

Step 1

we have to solve the given differential equations with intial condition.

We can solve it variable seperable

Step 2

Given$2ydx+3xdy=x{y}^{3}dy;x=1\text{}y=1$

$\Rightarrow 2ydx=x{y}^{3}dy-3xdy$

$\Rightarrow 2ydx=x({y}^{3}-3)dy$

$\Rightarrow \int \frac{2dx}{x}=\int \left(\frac{{y}^{3}-3}{y}\right)dy$

$\Rightarrow 2\int \frac{dx}{x}=\int ({y}^{2}-\frac{3}{y})dy$

$\Rightarrow 2\mathrm{ln}\left(x\right)=\frac{{y}^{3}}{3}-3\mathrm{ln}\left(y\right)+c$ (1)

Put$x=1,y=1$

$\Rightarrow 2\mathrm{ln}\left(1\right)=\frac{{1}^{3}}{3}-3\mathrm{ln}\left(1\right)+c$

$\Rightarrow 0=1-0+c$

$\therefore c=-1$

Put$c=-1\text{}\in \text{}\left(1\right)$

$2\mathrm{ln}\left(x\right)=\frac{{y}^{3}}{3}-3\mathrm{ln}\left(y\right)-1$

$\left(1\right)\text{}\int \frac{dx}{x}=\mathrm{ln}\left(x\right)+c\text{}\left(2\right)\text{}\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c$

we have to solve the given differential equations with intial condition.

We can solve it variable seperable

Step 2

Given

Put

Put

amarantha41

Expert

2021-12-21Added 38 answers

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

So,$2ydx+3xdy=x{y}^{3}dy$

$\Rightarrow 2y+3x\frac{dy}{dx}=x{y}^{3}\frac{dy}{dx}$

Substitute$\frac{dy}{dx}$ with y

So,

Substitute

RizerMix

Expert

2021-12-29Added 437 answers

Now using sepration of variable method then

Integrating both side then

Now, given initial condition x=1, y=1

Now by Eqn (1)

Put value of c in eq-n (1) then

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