Solving a 2nd Order Hyperbolic PDE with initial conditions

Question

I have the following hyperbolic PDE: $$u_{xx} + 5u_{xt} - 2u_{tt} = 0$$ with initial conditions:$$u(x,x) = sin(x)$$ $$u_x(x,x) = 0$$ My approach was to factor the equation, however even though I am aware that the general solution will be $$u(x,t) = f_1(t - 4x) + f_2(t+x)$$ I am having difficult incorporating the initial conditions to find a specific solution.

Answer 1

Sorry, the general solution should be $u(x,t)=f_1((5+\sqrt{33})x-2t)+f_2((5-\sqrt{33})x-2t)$ instead.

Sinilar to Solving Wave Equations with different Boundary Conditions :

$u(x,x)=\sin x$ :

$f_1((3+\sqrt{33})x)+f_2((3-\sqrt{33})x)=\sin x~......(1)$

$u_x(x,x)=0$ :

${f_1}_x((3+\sqrt{33})x)+{f_2}_x((3-\sqrt{33})x)=0$

$(3-\sqrt{33})f_1((3+\sqrt{33})x)+(3+\sqrt{33})f_2((3-\sqrt{33})x)=c~......(2)$

$\therefore f_1((3+\sqrt{33})x)=\dfrac{(3+\sqrt{33})\sin x-c}{2\sqrt{33}},f_2((3-\sqrt{33})x)=\dfrac{c-(3-\sqrt{33})\sin x}{2\sqrt{33}}$

$f_1(x)=-\dfrac{(3+\sqrt{33})\sin\dfrac{(3-\sqrt{33})x}{27}+c}{2\sqrt{33}},f_2(x)=\dfrac{(3-\sqrt{33})\sin\dfrac{(3+\sqrt{33})x}{27}+c}{2\sqrt{33}}$

$\therefore u(x,t)=\dfrac{(3-\sqrt{33})}{2\sqrt{33}}\sin\dfrac{(3+\sqrt{33})((5-\sqrt{33})x-2t)}{27}-\dfrac{(3+\sqrt{33})}{2\sqrt{33}}\sin\dfrac{(3-\sqrt{33})((5+\sqrt{33})x-2t)}{27}$