Asked by Mina
Solve the first-order initial value problem using half angle formulas
{y'(x)= 1 + sin^2 x
y (0)= -7
{y'(x)= 1 + sin^2 x
y (0)= -7
Answers
Answered by
Reiny
use the identity
cos 2x = 1 - 2sin^2 x
and solving for sin^2 x
sin^2 x = 1/2 - (1/2)cos 2x
then y'(x)= 1 + sin^2 x
becomes
y'(x)= 1 + sin^2 x
becomes
y'(x)= 1 + 1/2 - (1/2)cos 2x
= 3/2 - (1/2)cos 2x
this is now easy to integrate
Don't forget to add the constant.
plug in the point (0,-7) to find that constant.
cos 2x = 1 - 2sin^2 x
and solving for sin^2 x
sin^2 x = 1/2 - (1/2)cos 2x
then y'(x)= 1 + sin^2 x
becomes
y'(x)= 1 + sin^2 x
becomes
y'(x)= 1 + 1/2 - (1/2)cos 2x
= 3/2 - (1/2)cos 2x
this is now easy to integrate
Don't forget to add the constant.
plug in the point (0,-7) to find that constant.