Question

(7y=4t^2)dy+4tydt=0

Answer 1

$$(7y+4t^2)dy+4tydt=0$$

$$P(t, y)=7y+4t^2$$

$$Q(t, y)=4ty$$

$$\dfrac{\partial P}{\partial t}=8t$$

$$\dfrac{\partial Q}{\partial y}=4t$$

$$\dfrac{\partial P}{\partial t}\not=\dfrac{\partial Q}{\partial y}$$

Let

$$y(7y+4t^2)dy+4ty^2dt=0$$

$$M(t, y)=7y^2+4t^2y$$

$$N(t, y)=4ty^2$$

$$\dfrac{\partial M}{\partial t}=8ty$$

$$\dfrac{\partial N}{\partial y}=8ty$$

$$\dfrac{\partial M}{\partial t}=\dfrac{\partial N}{\partial y}$$

$$\dfrac{\partial F}{\partial y}=7y^2+4t^2y$$

$$F(y, t)=\dfrac{7}{3}y^3+2t^2y^2+\psi(t)=c_1$$

$$\dfrac{\partial F}{\partial t}=4ty^2+\dfrac{d\psi}{dt}=4ty^2=>\dfrac{d\psi}{dt}=0=>\psi=c_2$$

Then

$$\dfrac{7}{3}y^3+2t^2y^2=c$$