Question

Solve the equation in exact differentials:

(y-3x²)dx+(x-4y)dy=0

Answer 1

We shall solve the equation

$(y-3x^{2})dx +(x-4y)dy$

In exact differentials.

First of all, we shall show that the differential equation is indeed EXACT.

From the equation, we have

$M(x,y)=y-3x^{2}$ and

$N(x,y)=x-4y$

We know that for exactness,

$\frac{\partial M(x,y)}{\partial y}$ must be equal to $\frac{\partial N(x,y)}{\partial x}$

Now,

$\frac{\partial M(x,y)}{\partial y}=1$ and $\frac{\partial N(x,y)}{\partial x}=1$

Showing that the differential equation is EXACT.

We shall then solve the equation,

$F_{x}(x,y)=y-3x^{2}------(1)\\F_{y}(x,y)=x-4y------(2)$

Integrating equ(1) $w.r.t.$ $x$

$F(x,y)=xy-x^{3}+\phi(y)---(*)$

Differentiating $equ(*)$ $w.r.t.$ $y$

$F_{y}(x,y)=x+\phi^{\prime}(y)----(**)$

Equating $equ(2)$ and $equ(**)$

$x-4y=x+\phi^{\prime}(y)\\\implies\phi^{\prime}(y)=-4y\\$

Integrating both sides w.r.t $y$

$\phi(y)=-2y^{2}$

Now, from the equation

$F(x,y)=xy-x^{3}+\phi(y)$

Substituting the value of $\phi(y)=-2y^{2}$

We have

$F(x,y)=xy-x^{3}-2y^{2}$

Which is our solution to the differential equation.