Solve PDEs : $u_{xx}-10u_{xy}+25u_{yy}=e^x.$ Solve PDEs : $u_{xx}-10u_{xy}+25u_{yy}=e^x.$
Find solution of equation satisfy : $u(x,0)=e^x-\cos5x\\u_y(x,0)=\sin 5x. \quad(*)$
By set : $\begin{cases}\alpha=y+5x\\ \beta=y\end{cases}\Longrightarrow v_{\beta\beta}=\dfrac{e^{\frac{\alpha-\beta}{5}}}{25}$.
But I can't continue.
General solution form: $y.\varphi(y+5x)+\psi(y)+\frac{1}{25}\int e^{\frac{\alpha-\beta}{5}}$???????????????
 A: Let $\begin{cases}\alpha=5x+y\\\beta=5x-y\end{cases}$ ,
Then $u_x=u_\alpha\alpha_x+u_\beta\beta_x=5u_\alpha+5u_\beta$
$u_{xx}=(5u_\alpha+5u_\beta)_x=(5u_\alpha+5u_\beta)_\alpha\alpha_x+(5u_\alpha+5u_\beta)_\beta\beta_x=5(5u_{\alpha\alpha}+5u_{\alpha\beta})+5(5u_{\alpha\beta}+5u_{\beta\beta})=25u_{\alpha\alpha}+50u_{\alpha\beta}+25u_{\beta\beta}$
$u_y=u_\alpha\alpha_y+u_\beta\beta_y=u_\alpha-u_\beta$
$u_{xy}=(5u_\alpha+5u_\beta)_y=(5u_\alpha+5u_\beta)_\alpha\alpha_y+(5u_\alpha+5u_\beta)_\beta\beta_y=5u_{\alpha\alpha}+5u_{\alpha\beta}-(5u_{\alpha\beta}+5u_{\beta\beta})=5u_{\alpha\alpha}-5u_{\beta\beta}$
$u_{yy}=(u_\alpha-u_\beta)_y=(u_\alpha-u_\beta)_\alpha\alpha_y+(u_\alpha-u_\beta)_\beta\beta_y=u_{\alpha\alpha}-u_{\alpha\beta}-(u_{\alpha\beta}-u_{\beta\beta})=u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta}$
$\therefore25u_{\alpha\alpha}+50u_{\alpha\beta}+25u_{\beta\beta}-10(5u_{\alpha\alpha}-5u_{\beta\beta})+25(u_{\alpha\alpha}-2u_{\alpha\beta}+u_{\beta\beta})=e^\frac{\alpha+\beta}{10}$
$100u_{\beta\beta}=e^\frac{\alpha+\beta}{10}$
$u_{\beta\beta}=\dfrac{e^\frac{\alpha+\beta}{10}}{100}$
$u_\beta=p(\alpha)+\dfrac{e^\frac{\alpha+\beta}{10}}{10}$
$u(\alpha,\beta)=\beta p(\alpha)+q(\alpha)+e^\frac{\alpha+\beta}{10}$
$u(x,y)=(5x-y)p(5x+y)+q(5x+y)+e^x$
$u(x,y)=-2yp(5x+y)+(5x+y)p(5x+y)+q(5x+y)+e^x$
$u(x,y)=yf\left(x+\dfrac{y}{5}\right)+g\left(x+\dfrac{y}{5}\right)+e^x$
$u(x,0)=e^x-\cos5x$ :
$g(x)+e^x=e^x-\cos5x$
$g(x)=-\cos5x$
$\therefore u(x,y)=yf\left(x+\dfrac{y}{5}\right)-\cos(5x+y)+e^x$
$u_y(x,0)=\sin5x$ :
$f(x)+\sin5x=\sin5x$
$f(x)=0$
$\therefore u(x,y)=e^x-\cos(5x+y)$
