Answer to Question #127090 in Differential Equations for koshila

Question

(2t×siny+e to the power t×y to the power 3)dt+(t to the power 2×cosy+3×y to the power 2×e to the power t)dy=0

Answer 1

D.E is

$$(2t\sin(y)+e^ty^3)dt+(t^2\cos(y)+3y^2e^t)dy=0$$

Now .divide both side of the above equation by $dt$ ,thus we get

$$(2t\sin(y)+e^ty^3)+(t^2\cos(y)+3y^2e^t)\frac{dy}{dt}=0\\ \implies 2t\sin(y)+t^2\cos(y)\frac{dy}{dt}+e^ty^3+3y^2e^t\frac{dy}{dt}=0\\$$

Note that

$$2t\sin(y)+t^2\cos(y)\frac{dy}{dt}=\frac{d}{dt}(t^2\sin(y))\\ e^ty^3+3y^2e^t\frac{dy}{dt}=\frac{d}{dt}(y^3e^t)$$

Thus,

$$2t\sin(y)+t^2\cos(y)\frac{dy}{dt}+e^ty^3+3y^2e^t\frac{dy}{dt}=0\\\implies \frac{d}{dt}(t^2\sin(y)+y^3e^t)=0\\ \implies \int d(t^2\sin(y)+y^3e^t)=\int0dt\\ \implies t^2\sin(y)+y^3e^t=c$$