I can't figure out what method to use for this question. I tried the coordinate method to find the equation, but that didn't work. Then I tried the Lagrange's auxiliary equations method, but that didn't work either. I'm just trying to find the appropriate method to solve the above PDE.
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You can solve it using the same method used to solve linear inhomogeneous ODEs: write the solution as $u=v+w$, where $v$ is the general solution to the homogeneous PDE, $$ 2v_x-v_y=10v, \tag{1} $$ and $w$ is a particular solution to the inhomogeneous PDE, $$ 2w_x-w_y=10w+5e^{x-3y}. \tag{2} $$ By inspection, it's easy to guess that Eq. $(2)$ has a particular solution of the form $w(x,y)=Ae^{x-3y}$ (I leave it to you to determine $A$). It's also simple to find the general solution to Eq. $(1)$ by solving the Lagrange-Charpit equations, $$ \frac{dx}{2}=\frac{dy}{-1}=\frac{dv}{10v}. \tag{3} $$ It's so simple, in fact, that I'll just write the result: $v(x,y)=e^{5x}f(x+2y)$, where $f$ is an arbitrary differentiable function. Therefore, the general solution to the original PDE is $$ u(x,y)=v(x,y)+w(x,y)=e^{5x}f(x+2y)+Ae^{x-3y}. \tag{4} $$ Finally, use the boundary condition $u(x,0)=x^2e^{4x}-e^x$ to determine $f$.
Gonçalo
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thanks! For eq 2, how did you guess the particular solution is of that form? We just started learning pdes, so I'm not quite sure. – starry41 Jan 20 '24 at 01:07
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The exponential function has this property: $\frac{d}{dz}e^{cz}=ce^{cz}$. – Gonçalo Jan 20 '24 at 01:15
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oh yes I do know that, but I'm confused as to how you solved the entire equation. Like there were more parts to it, how did you differentiate $e^{x-3y}$ only? – starry41 Jan 20 '24 at 01:19
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Are you familiar with linear ODEs? With the method of undetermined coefficients? – Gonçalo Jan 20 '24 at 01:37
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yes I am. I just tried it on equation 2 and got A=-1, is that right? – starry41 Jan 20 '24 at 01:55
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1Yes, that's correct. – Gonçalo Jan 20 '24 at 01:57