Answer to Question #205543 in Differential Equations for Devansh

Question

(D2-3DD'+2D'2)=e2x-y+ex+y

Answer 1

Answer:-

$$\left(D^2-3DD'+2D'^2\right)z=e^{2x-y}+e^{x+y}$$

This is an inhomogeneous equation, so its solution consists of two parts

$$z(x,y)=(C.F.)+(P.I.)\\[0.3cm] (C.F.)-\text{Complementary Function of Homogeneous Linear PDE}\\[0.3cm] (P.I.)-\text{Particular Integral}$$

1 STEP : We try to find $(C.F.)$

As we know, the equation

$$\boxed{\left(D-mD'\right)z=0\to z(x,y)=\phi(y+mx)}\\[0.3cm] \phi(y+mx)\,\text{is an arbitrary function}$$

In our case,

$$\left(D^2-3DD'+2D'^2\right)z=\left(D^2-DD'-2DD'+2D'^2\right)z=\\[0.3cm] =\left(D\left(D-D'\right)-2D'\left(D-D'\right)\right)z=\\[0.3cm] =\left(D-D'\right)\left(D-2D'\right)z\\[0.3cm] \left(D-D'\right)\left(D-2D'\right)z=0\longrightarrow\\[0.3cm] \left\{\begin{array}{l} \left(D-1D'\right)z=0\to z(x,y)=\phi_1(y+x)\\[0.3cm] \left(D-2D'\right)z=0\to z(x,y)=\phi_2(y+2x) \end{array}\right.$$

Conclusion,

$$\boxed{(C.F.)=\phi_1(y+x)+\phi_2(y+2x)}\\[0.3cm] \phi_1,\phi_2\,\text{are arbitrary functions}$$

2 STEP : We try to find $(P.I.)$

As we know

$$f\left(D,D'\right)z=F(x,y)\to(P.I.)=\frac{1}{f\left(D,D'\right)}F(x,y)\\[0.3cm] \text{special case :}\,\,\,\frac{1}{f\left(D,D'\right)}e^{ax+by}=\frac{e^{ax+by}}{f(a,b)},\,\,\,f(a,b)\neq0\\[0.3cm] if\,\,\,f(a,b)=0 : (P.I.)=x\cdot\frac{1}{f'\left(D,D'\right)}\cdot e^{ax+by},\\[0.3cm] f'\left(D,D'\right)\text{ is the partial derivative of}\,\,\,f\left(D,D'\right)\,\,\,\text{ w.r.t D}$$

In our case,

$$\left(D^2-3DD'+2D'^2\right)z=\underbrace{e^{2x-y}}_{(P.I.)_1}+\underbrace{e^{x+y}}_{(P.I.)_2}\\[0.3cm] (P.I.)_1 : \left(D^2-3DD'+2D'^2\right)z=e^{2x-y}\to\\[0.3cm] (P.I.)_1=\frac{1}{\left(D^2-3DD'+2D'^2\right)}\cdot e^{2x-y}=\frac{e^{2x-y}}{2^2-3\cdot2\cdot(-1)+2(-1)^2}=\\[0.3cm] =\frac{e^{2x-y}}{4+6+2}=\frac{e^{2x-y}}{12}\to\boxed{(P.I.)_1=\frac{e^{2x-y}}{12}}\\[0.3cm] (P.I.)_2=\frac{1}{\left(D^2-3DD'+2D'^2\right)}\cdot e^{x+y}=x\cdot\frac{1}{2D-3D'}\cdot e^{x+y}=\\[0.3cm] =\frac{x\cdot e^{x+y}}{2-3}=-x\cdot e^{x+y}\to\boxed{(P.I.)_2=-x\cdot e^{x+y}}$$

General conclusion.

$$\left(D^2-3DD'+2D'^2\right)z=e^{2x-y}+e^{x+y}\to\\[0.3cm] z(x,y)=(C.F.)+(P.I.)_1+(P.I.)_2\\[0.3cm] \boxed{z(x,y)=\phi_1(y+x)+\phi_2(y+2x)+\frac{e^{2x-y}}{12}-x\cdot e^{x+y}}\\[0.3cm] \phi_1,\phi_2\,\text{are arbitrary functions}$$

ANSWER

$$z(x,y)=\phi_1(y+x)+\phi_2(y+2x)+\frac{e^{2x-y}}{12}-x\cdot e^{x+y}\\[0.3cm] \phi_1,\phi_2\,\text{are arbitrary functions}$$