Question

.(6𝑥2 −𝑦+3)𝑑𝑥+(3𝑦2 −𝑥−2)𝑑𝑦=0

Answer 1

Ans:-

$.(6𝑥^2 −𝑦+3)𝑑𝑥+(3𝑦^2 −𝑥−2)𝑑𝑦=0$

We consider here the following standard form of ordinary differential equation: $P(x, y)dx + Q(x, y)dy = 0$  

$P(x,y)=6x^2-y+3$ and $Q(x,y)=3y^2-x-2$

And we know that $\dfrac{\delta P}{\delta y}=\dfrac{\delta Q}{\delta x}$ then

solution of differential equation will be

$\Rightarrow$ $du =\dfrac{ ∂u} {∂x} dx +\dfrac {∂u} {∂y} dy$

$\Rightarrow P dx + Q dy = 0 .$

One solves $\dfrac{∂u }{∂x} = P$ and $\dfrac{∂u}{ ∂y} = Q$ to find $u(x, y)$

Now

$\dfrac{∂u}{ ∂y} = 3y^2-x-2\\ u=y^3-(x+2)y+c,$ $\dfrac{∂u }{∂x} = P\\ u=2x^3-(y-3)x+c$

Then $du = 0$ gives $u(x, y) = C,$ where C is a constant.

$\Rightarrow u(x,y)=2x^3+y^3-(x+2)y-(y-3)x+C$

It will be final solution of the differential equation.