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 \(2y-1=e^x(sinx-cosx).\)