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Derive the solution of the problem $$u_{xy} + (\tan(y))u_x = 2 x \tan(y)$$ with $u(x,0) = x^2 + e^{x^3}$ and $u(0,y) = y^{10} + \cos(y)$.


How do I solve this question?, I think it's by separation of variables, but I can't figure it out.

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Multiply both sides by $1/ \cos y$, integrate with respect to $x$ and write: \begin{equation} \dfrac{\partial}{\partial y} \left( \dfrac{u-x^2}{\cos y} \right) = f(y) \end{equation} So we integrate this once again and write: \begin{equation} u=x^2 + F(y) \cos y+G(x)\cos y. \end{equation} Apply boundary conditions to get: \begin{equation} u(x,y) = x^2 + e^{x^3} \cos y + y^{10}. \end{equation}

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Basically the same approach with some more details. You could notice that you equation does not contain $u$ but only partial derivatives. Set $v(x,y)=u_x$, which should solve $$v_y+(\tan y) v=2x\tan y.$$ Note that this is an ODE in $y$ at fixed $x$. Find the solution, first the homogeneous equation $v(x,y)=\lambda(x,y)\cos y$ that you plug into the full equation to get $\lambda(x,y)=\frac{2x}{\cos y}+p(x)$ for an arbitrary $p$, so that you get $$v(x,y)=2x+p(x)\cos y$$ Coming back to $u$ by integrating in $x$, you get $$u(x,y)=x^2+P(x)\cos y +Q(y)$$ where $P,Q$ are arbitrary functions. You will set these using boundary conditions: $u(x,0)=x^2+P(x)=x^2+e^{x^3}+Q(0)$ give $P(x)+Q(0)=e^{x^3}$ and then $u(0,y)=(1-Q(0))\cos y+Q(y)=y^{10}+\cos y$ provides $Q(y)-Q(0)\cos y=y^{10}$. Leading to $$u(x,y)=x^2+e^{x^3}\cos y+y^{10}.$$

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