Question

Find the equation of a curve passing through the point (0, 0) and whose differential equation is .

Answer 1

The differential equation of the curve is given below,

y ′ = e x sinx dy= e x sinxdx

Integrate both sides.

y= ∫ e x sinxdx

Use integration by parts and consider sinxas first function and e x as second function.

I= ∫ e x sinxdx I=sinx ∫ e x dx − ∫ ( d dx sinx ∫ e x dx )dx I= e x sinx− ∫ e x cosxdx

Again use integration by parts and consider cosx as first function and e x as second function.

I= e x sinx−[ cosx ∫ e x dx − ∫ ( d dx cosx ∫ e x dx )dx ] I= e x sinx− e x cosx+( − ∫ e x sinxdx ) I= e x ( sinx−cosx )−I

Further simplify.

2I= e x ( sinx−cosx )+C I= e x 2 ( sinx−cosx )+C

So,

y= e x 2 ( sinx−cosx )+C(1)

Now, this curve passes through point ( 0,0 ) so substitute values for x and y.

0= e 0 ( sin0−cos0 ) 2 +C C=− −1 2 C= 1 2

Substitute the values into equation (1).

y= e x 2 ( sinx−cosx )+ 1 2 2y= e x ( sinx−cosx )+1 2y−1= e x ( sinx−cosx )

Thus, the required equation of the curve passing through ( 0,0 ) is 2y−1= e x ( sinx−cosx ).