Answer to Question #204996 in Calculus for nowrin sultana

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

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