#### 2.1.77 $$u_x u_y = 1$$

problem number 77

From UMN Math 5587 HW2, Fall 2016, problem 5(a).

Solve for $$u(x,y)$$ \begin {align*} u_x u_y = 1 \end {align*}

Mathematica

$\left \{\left \{u(x,y)\to \frac {x}{c_2}+c_2 y+c_1\right \}\right \}$

Maple

$u \left (x , y\right ) = y \mathit {\_c}_{2}+c_{1}+c_{2}+\frac {x}{\mathit {\_c}_{2}}$

Hand solution

Solve $$u_{x}u_{y}=1$$. Let $$u=f\left ( x\right ) +g\left ( y\right )$$. The pde becomes $$f^{\prime }\left ( x\right ) g^{\prime }\left ( y\right ) =1$$ or $$f^{\prime }\left ( x\right ) =\frac {1}{g^{\prime }\left ( y\right ) }$$. Hence both are constant, say $$\lambda$$. Therefore $$f^{\prime }\left ( x\right ) =\lambda$$ or $$f\left ( x\right ) =\lambda x+C_{1}$$ and similarly, $$\frac {1}{g^{\prime }\left ( y\right ) }=\lambda$$ or $$g^{\prime }\left ( y\right ) =\frac {1}{\lambda }$$ or $$g\left ( y\right ) =\frac {y}{\lambda }+C_{2}$$. Therefore the solution becomes\begin {align*} u\left ( x,y\right ) & =f\left ( x\right ) +g\left ( y\right ) \\ & =\lambda x+C_{1}+\frac {y}{\lambda }+C_{2} \end {align*}

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