Question

Find the equation of a curve passing through the point $(0,0)$ and whose differential equations is $y'=e^x sinx.$

Answer 1

The differential equation of the curve is:
$y'=e^x sinx.$
$⇒\frac{dy}{dx}=e^x sinx$
$⇒dy=e^x sinx\,dx$
Integrating both sides,we get:
$∫dy=∫e^x sinx\, dx...(1)$
Let $I=∫e^x sinx\, dx.$
⇒$I=sinx ∫e^x dx-∫(\frac{d}{dx}(sinx).∫e^x dx)dx$
$⇒I=sinx.e^x-∫cosx.e^x dx$
$⇒I=sinx.e^x-[cosx.∫e^x dx-∫(\frac{d}{dx}(cosx).∫e^x dx)dx]$
$⇒I=sinx.e^x-[cosx.e^x-∫(-sinx).e^x dx]$
$⇒I=e^x sinx-e^x cosx-I$
$⇒2I=e^x(sinx-cosx)$
$⇒I=\frac{e^x(sinx-cosx)}{2}$
Substituting this value in equation(1),we get:
$y=\frac{e^x(sinx-cosx)}{2}+C...(2)$
Now,the curve passes through point(0,0).
$∴0=\frac{e^0(sin0-cos0)}{2}+C$
$⇒0=\frac{1(0-1)}{2}+C$
$⇒C=\frac{1}{2}$
Substituting $C=\frac{1}{2}$ in equation(2),we get:
$y=\frac{e^x(sinx-cosx)}{2}+\frac{1}{2}$
$⇒2y=e^x(sinx-cosx)+1$
$⇒2y-1=e^x(sinx-cosx)$
Hence,the required equation of the curve is $2y-1=e^x(sinx-cosx).$