Find the initial value problem

Question

Can u help me solve this problem:

 (cos y) dy/dx - sin(y) = x, where y(0)=0

 

 

 

Answer 1

Hi Vo,

        I'll try and get you started first we can put the above equation in the following form which will help when we try and find a solution.

 

                            (x+sin(y))dx -cos(y)dy=0

 

Now there is a class of differential equations that are called exact these equations are of the form

 

                            M(x,y)dx+N(x,y)dy=0 now if ∂M/∂y=∂N/∂x we can find a function g(x,y)=c that will be a potential solution.

                            for our problem M(x,y)=x+sin(y) and N(x,y)=-cos(y) and the partials are cos(y) and 0 so they are not equal.

                            In these cases we can look for an "integrating factor" u(x) such that 

                    

                            u(x)M(x,y) +u(x)N(x,y)dy/dx =0 is exact, This means that 

 

                             ∂(u(x)M(x,y))/∂y=∂(u(x)N(x,y))∂x carrying out the differentials gives

 

                             - cos(y)u(x)=cos(y)du/dx this turns out to be du/dx/u=-1 integrating both sides gives ln(u)=-x

                             

                               so u(x)=e^-x is the integrating factor. Now u(x)M(x,y) and u(x,y)N(x,y) form a differential equation t

                             

                            is exact because ∂(u(x)M(x,y)))/∂y=e^-xcos(y) and ∂(u(x)N(x,y)))/∂x=e^-xcos(y)

 

                            So the solution is ∫e^-x(x+sin(y₀))dx +∫e^-xcos(y)dy first integral limits are 0 to x and the second is 0 

                            to y.

 

      Hopefully you've had enough differential equation theory to follow my notation.

Regards

Jim