Y'' plus y equals cotx diff eq?

Answer 1

In order to solve this inhomogeneous differential equation you need to start by solving the homogeneous case first (aka when the right hand side is just 0).

The characteristic equation for this differential equation is r²+1=0 or r²=-1 which means that r must be equal to ±i. Therefore, the general solution to this homogeneous problem Is y=c1*sin(x)+c2*cos(x) where c1 and c2 are constants determined by the initial conditions.

In order to solve the inhomogeneous problem we need to first find the Wronskian of our two solutions.

_________|y1(x) y2(x) | __| sin(x) cos(x) |

W(y1, y2)= |y1'(x) y2'(x) | = | cos(x) -sin(x) | = -sin(x)²-cos(x)²= -1

Next, we calculate the particular solution

Y(x)=-sin(x)* Integral(-1*cos(x)*cot(x)) + cos(x)*Integral(-1*sin(x)*cot(x))

=sin(x)*Integral(cos²(x)/sin(x)) - cos*Integral(cos(x))

=sin(x)*(ln(tan(x/2)) + cos(x)) -cos(x)*sin(x)=sin(x)*ln(tan(x/2))

Finally, to answer the entire equation, we add the particular solution to the homogeneous solution to get

y(x)=sin(x)*ln(tan(x/2)) + c1*sin(x)+c2*cos(x)