Question

Cos ( x + y ) p +sin ( x + y ) q,=z

Answer 1

Answer:

The given equation:

cos(x+y)∂z∂x+sin(x+y)∂z∂y=z(1)

We solve this P.D.E by using the method of characteristics. Consider the characteristic equation of equations (1) as follows:

dxcos(x+y)=dysin(x+y)=dzz

Rearrange these equations to have a system of ODEs as below:

y′=dydx=tan(x+y)(2)

And

dzz=d(x+y)sin(x+y)+cos(x+y)(3)

We first solve equation (2). To do so, let u=x+y⇒u′=1+y′⇒y′=u′−1 then equation (2) becomes:

u′=1+tanu

or

dx=du1+tanu

Now let v=tanu⇒dv=(1+tan2u)du=(1+v2)du. Substitute this into equation above to attain:

dx=dv(1+v)(1+v2)

or

dx=(12(1+v)+12(1+v2)−v2(1+v2))dv

Integrating this obtains:

x+C1=12ln|1+v|−14ln(1+v2)+12artanv

where C1 is an integration constant.