Question

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

Answer 1

Answer:

$y =\frac{e^{x}}{2}*[sin x -cos x]$ + 1/2,

Step-by-step explanation:

Hi,

Given differential equation is $y' = e^{x}sin x$,

which can be written as

$\frac{dy}{dx} = e^{x}sin x$,

Separating out x terms to one side and y terms to other,

we get

$dy = e^{x}sin x dx$,

Now, integrating on both sides, we get

$\int dy = \int e^{x}sin x dx$,

$y = \int e^{x}sin x dx$,

$Let I = \int e^{x}sin x dx$,

We get $y = I$,-----(1)

$I = \int e^{x} sin x - \int((\int e^{x})*(sin x)')$,

$I = e^{x}sin x - \int e^{x})*cos x$,

Let $I' = \int e^{x}cos x$,

$I = e^{x} sin x - I'$,-----(2)

$I' =cos x (\int e^{x})- \int(\int e^{x}dx)*(cos x)'dx$,

$= e^{x}cos x - \int e^{x}*(-sin x)dx$,

$I'=e^{x} cos x + I$,

Substituting value of I' in equation (2), we get

$I =e^{x} sin x -(e^{x} cos x + I)$,

$2I =e^{x} sin x -e^{x} cos x$,

$I =\frac{1}{2}*[e^{x} sin x -e^{x} cos x]$,

$I =\frac{e^{x}}{2}*[sin x -cos x]$,

Substituting the value of I in (1), we get

$y =\frac{e^{x}}{2}*[sin x -cos x]$ + c,

Given that, the above curve passes through (0, 0),

By substituting, we get

0 = 1/2(-1) + c

c = 1/2

So, the particular solution will become

$y =\frac{e^{x}}{2}*[sin x -cos x]$ + 1/2,

Hope, it helps !