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

Solve the PDE by method of separating variables (∂^2 u/∂x^2 ) + 2t(∂^2 u/(∂x∂t)) − 4u = 0

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

Commented by Joel578 last updated on 16/Feb/20

My approach Let the solution be u(x,t) = F(x)G(t) Substitute u(x,t) to original eq. and divide by F(x)G(t) yields (1/F) (d^2 F/dx^2 ) + 2 ((1/F) (dF/dx))((t/T) (dT/dt)) − 4 = 0 Let P(x) = (1/F) (d^2 F/dx^2 ), M(x) = (2/F) (dF/dx) , N(t) = (t/T) (dT/dt) ⇒ P(x) + M(x)N(t) − 4 = 0 Differentiate both sides with respect to x, then t ⇒ M ′(x) N ′(t) = 0 which means either M(x) or N(t) is a constant • Case 1 : N(t) = n P(x) + n M(x) − 4 = 0 ⇒ (d^2 F/dx^2 ) + 2n (dF/dx) − 4F = 0 ⇒ F(x) = C_1 e^((−n + 2(√(4 + n^2 )))x) + C_2 e^((−n − 2(√(4 + n^2 )))x) ⇒ N(t) = (t/T) (dT/dt) = n ⇒ T(t) = C_3 t^n • Case 2 : M(x) = m P(x) + m N(t) − 4 = 0

$$\mathrm{My}\:\mathrm{approach} \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{be}\:\:{u}\left({x},{t}\right)\:=\:{F}\left({x}\right){G}\left({t}\right) \\ $$$$\mathrm{Substitute}\:{u}\left({x},{t}\right)\:\mathrm{to}\:\mathrm{original}\:\mathrm{eq}.\:\mathrm{and}\:\mathrm{divide}\: \\ $$$$\mathrm{by}\:{F}\left({x}\right){G}\left({t}\right)\:\mathrm{yields} \\ $$$$\frac{\mathrm{1}}{{F}}\:\frac{{d}^{\mathrm{2}} {F}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\:\left(\frac{\mathrm{1}}{{F}}\:\frac{{dF}}{{dx}}\right)\left(\frac{{t}}{{T}}\:\frac{{dT}}{{dt}}\right)\:−\:\mathrm{4}\:=\:\mathrm{0}\: \\ $$$$ \\ $$$$\mathrm{Let}\:{P}\left({x}\right)\:=\:\frac{\mathrm{1}}{{F}}\:\frac{{d}^{\mathrm{2}} {F}}{{dx}^{\mathrm{2}} },\:\:\:{M}\left({x}\right)\:=\:\frac{\mathrm{2}}{{F}}\:\frac{{dF}}{{dx}}\:,\:\:\:{N}\left({t}\right)\:=\:\frac{{t}}{{T}}\:\frac{{dT}}{{dt}} \\ $$$$\Rightarrow\:{P}\left({x}\right)\:+\:{M}\left({x}\right){N}\left({t}\right)\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{Differentiate}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:{x},\:\mathrm{then}\:{t} \\ $$$$\Rightarrow\:{M}\:'\left({x}\right)\:{N}\:'\left({t}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{means}\:\mathrm{either}\:{M}\left({x}\right)\:\mathrm{or}\:{N}\left({t}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant} \\ $$$$ \\ $$$$\bullet\:\mathrm{Case}\:\mathrm{1}\::\:{N}\left({t}\right)\:=\:{n} \\ $$$${P}\left({x}\right)\:+\:{n}\:{M}\left({x}\right)\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\frac{{d}^{\mathrm{2}} {F}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}{n}\:\frac{{dF}}{{dx}}\:−\:\mathrm{4}{F}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{F}\left({x}\right)\:=\:{C}_{\mathrm{1}} {e}^{\left(−{n}\:+\:\mathrm{2}\sqrt{\mathrm{4}\:+\:{n}^{\mathrm{2}} }\right){x}} \:+\:{C}_{\mathrm{2}} \:{e}^{\left(−{n}\:−\:\mathrm{2}\sqrt{\mathrm{4}\:+\:{n}^{\mathrm{2}} }\right){x}} \\ $$$$\Rightarrow\:{N}\left({t}\right)\:=\:\frac{{t}}{{T}}\:\frac{{dT}}{{dt}}\:=\:{n}\:\Rightarrow\:{T}\left({t}\right)\:=\:{C}_{\mathrm{3}} \:{t}^{{n}} \\ $$$$ \\ $$$$\bullet\:\mathrm{Case}\:\mathrm{2}\::\:{M}\left({x}\right)\:=\:{m} \\ $$$${P}\left({x}\right)\:+\:{m}\:{N}\left({t}\right)\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$

Commented by Joel578 last updated on 16/Feb/20

Please guide me with the second case.

$${Please}\:{guide}\:{me}\:{with}\:{the}\:{second}\:{case}. \\ $$

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