Question

ydx-xdy-3x²y²e^x³dx=0​

Answer 1

Answer:

$\large\bold\red{ c = \frac{x}{y} - {e}^{ {x}^{3} } }$

Step-by-step explanation:

Given,

A differential equation,

• $ydx - xdy - 3 {x}^{2} {y}^{2} {e}^{ {x}^{3} } dx = 0$

To solve this DE,

Divide both sides with $\bold{{y}^{2}}$

Therefore,

We get,

$= > \frac{ydx - xdy - 3 {x}^{2} {y}^{2} {e}^{ {x}^{3} } dx }{ {y}^{2} } = 0 \\ \\ = > \frac{ydx - xdy}{ {y}^{2} } = \frac{ 3{x}^{2} {y}^{2} {e}^{ {x}^{3} } dx}{ {y}^{2} } \\ \\ = > d( \frac{x}{y} ) = 3{x}^{2} {e}^{ {x}^{3} } dx$

Now,

Integrating both the sides,

We get,

$= > \int \: d( \frac{x}{y} ) = \int3 {x}^{2} {e}^{ {x}^{3} } dx \\ \\ = > \frac{x}{y} = \int3 {x}^{2} {e}^{ {x}^{3} } dx \: \: \: \: \: \: .........(i)$

Now,

Let us assume that

${x}^{3} = t$

Differentiating both the sides,

We get,

$= > 3 {x}^{2} dx = dt$

Now,

Substituting this value in eqn $(i)$,

We get,

$= > \frac{x}{y} =\int {e}^{t} dt \\ \\ = > \frac{x}{y} = {e}^{t} + c$

Substituting the value of t,

We get,

$= > \frac{x}{y} = {e}^{ {x}^{3} } + c \\ \\ = > \bold{ c = \frac{x}{y} - {e}^{ {x}^{3} } }$