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Question Number 81937 by Joel578 last updated on 16/Feb/20

$$\mathrm{Solve}\mathrm{the}\mathrm{PDE}\mathrm{by}\mathrm{method}\mathrm{of}\mathrm{separating}\mathrm{variables}$$$$\frac{{\partial }^{2}u}{\partial x^2}+2t\frac{{\partial }^{2}u}{\partial x\partial t}-4u=0$$

Commented by Joel578 last updated on 16/Feb/20

$$\mathrm{My}\mathrm{approach}$$$$\mathrm{Let}\mathrm{the}\mathrm{solution}\mathrm{be}u(x,t)=F\left(x\right)G\left(t\right)$$$$\mathrm{Substitute}u(x,t)\mathrm{to}\mathrm{original}\mathrm{eq}.\mathrm{and}\mathrm{divide}$$$$\mathrm{by}F\left(x\right)G\left(t\right)\mathrm{yields}$$$$\frac{1}{F}\frac{d^{2}F}{{dx}^2}+2\left(\frac{1}{F}\frac{dF}{dx}\right)\left(\frac{t}{T}\frac{dT}{dt}\right)-4=0$$$$$$$$\mathrm{Let}P\left(x\right)=\frac{1}{F}\frac{d^{2}F}{{dx}^2},M\left(x\right)=\frac{2}{F}\frac{dF}{dx},N\left(t\right)=\frac{t}{T}\frac{dT}{dt}$$$$\Rightarrow P\left(x\right)+M\left(x\right)N\left(t\right)-4=0$$$$\mathrm{Differentiate}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{then}t$$$$\Rightarrow M{}^{\prime }\left(x\right)N{}^{\prime }\left(t\right)=0$$$$\mathrm{which}\mathrm{means}\mathrm{either}M\left(x\right)\mathrm{or}N\left(t\right)\mathrm{is}\mathrm{a}\mathrm{constant}$$$$$$$$\bullet \mathrm{Case}1:N\left(t\right)=n$$$$P\left(x\right)+nM\left(x\right)-4=0$$$$\Rightarrow \frac{d^{2}F}{{dx}^2}+2n\frac{dF}{dx}-4F=0$$$$\Rightarrow F\left(x\right)=C_1{e}^{(-n+2\sqrt{4+n^2})x}+C_2{e}^{(-n-2\sqrt{4+n^2})x}$$$$\Rightarrow N\left(t\right)=\frac{t}{T}\frac{dT}{dt}=n\Rightarrow T\left(t\right)=C_3{t}^n$$$$$$$$\bullet \mathrm{Case}2:M\left(x\right)=m$$$$P\left(x\right)+mN\left(t\right)-4=0$$

Commented by Joel578 last updated on 16/Feb/20

$$Pleaseguidemewiththesecondcase.$$

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