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To Solve: $\displaystyle z=px+qy+sin(x+y)$, where $\displaystyle p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}$

As per theory, there are four ways to solve a non-linear PDE of first order..

i) f(p,q)=0 ii)f(z,p,q)=0 iii)f(x,p)=F(y,q) iv) z=px+qy+f(p,q)

This does not seem to fit in the above four ways directly. I feel that some transformation is required, but what ?

Please advise.

The given answer is : $\displaystyle z=ax+by+sin(a+b) $

The answer makes me feel that its a type (i) problem.Am I right?

square_one
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In fact this is just a first-order linear PDE. You just need to solve it by method of characteristics.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$

$\dfrac{dz}{dt}=z-\sin(x+y)=z-\sin((y_0+1)e^t)$ , letting $z(0)=f(y_0)$ , we have $z=e^tf(y_0)-e^t\int_0^te^{-t}\sin((y_0+1)e^t)~dt=xf\left(\dfrac{y}{x}\right)-x\int_0^{\ln x}e^{-t}\sin\left(\left(\dfrac{y}{x}+1\right)e^t\right)dt$

doraemonpaul
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