Find the equation of a curve passing through the point (0, 0) and whose differentialequation is `y^(prime)=e xsinx`

Explanation

Given, differential equation of curve,
`y'=e^(x)sinx` or `(dy)/(dx)=e^(x)sinx`
`implies dy=e^(x)sinxdx`
`implies intdy=inte^(x)sinxdx`………`(1)`
Let `I=inte^(x)sinxdx`………..`(2)`
`implies I=sinxinte^(x)dx-int[(d)/(dx)(sinx)inte^(x)dx]dx`
`=sinxinte^(x)dx-intcosx*e^(x)dx`
`implies = sinx*e^(x)-[cosx*e^(x)dx-int((d)/(dx)(cosx)*inte^(x)dx)dx]`
`I=sinx*e^(x)-[cosx*e^(x)-int(-sinx)*e^(x)dx]`
`=sinx*e^(x)-e^(x)*cosx-inte^(x)*sinx*dx`
`implies I=sinx*e^(x)-cos*e^(x)-I` [from equation `(2)`]
`implies 2I=e^(x)(sinx-cosx)`
`implies I=(e^(x)(sinx-cos))/(2)`
put this value in equation `(1)`,
`y=(e^(x)(sinx-cosx))/(2)+C`...........`(3)`
Curve passes through the point `(0,0)`. Therefore put `x=0`, `y=0`
`:. 0=(e^(0)(sin0-cos0))/(2)+C` [from equation `(3)`]
`0=(1(0-1))/(2)+CimpliesC=(1)/(2)`
put `C=(1)/(2)` in equation `(3)`,
`y=(e^(x)(sinx-cosx))/(2)+(1)/(2)`
`implies 2y=e^(x)(sinx-cosx)+1`
`implies 2y-1=e^(x)(sinx-cosx)`
which is the equation of required curve.