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- Homework Statement
- This is a question set by myself;

Find the general solution of

##-4u_x+6u_y+10u=e^{x+2y}##

- Relevant Equations
- Method of characteristic.

My take;

##ξ=-4x+6y## and ##η=6x+4y##

it follows that,

##52u_ξ +10u=e^{x+2y}##

for the homogenous part; we shall have the general solution;

$$u_h=e^{\frac{-5}{26} ξ} f{η }$$

now we note that

$$e^{x+2y}=e^{\frac{8ξ+η}{26}}$$

that is from solving the simultaneous equation;

##ξ=-4x+6y## and ##η=6x+4y## where ##x=\dfrac{12ξ-18η}{156}## and ##y=\dfrac{3ξ+2η}{26}##

for the inhomogenous part, we shall have;

$$u_p = e^{\frac{-5}{26}ξ}\int \frac{e^{\frac{8ξ+η}{26}}⋅e^{\frac{5}{26}ξ}}{52} dξ$$$$u_p = \frac{e^{\frac{13ξ+η}{26}}⋅e^{\frac{-5}{26}ξ}}{26} $$

$$u(ξ,η)=u_h+u_p=e^{\frac{-5}{26} ξ} \left[ f(η)+\frac{e^{\frac{13ξ+η}{26}}}{26}\right]$$

$$u(x,y)=

e^{\frac{-5}{26} (-4x+6y)} \left[ f(6x+4y)+\frac{1}{26} {e^{\frac{-46x+82y}{26}}}\right]$$your insight is welcome or alternative approach to this problem.

##ξ=-4x+6y## and ##η=6x+4y##

it follows that,

##52u_ξ +10u=e^{x+2y}##

for the homogenous part; we shall have the general solution;

$$u_h=e^{\frac{-5}{26} ξ} f{η }$$

now we note that

$$e^{x+2y}=e^{\frac{8ξ+η}{26}}$$

that is from solving the simultaneous equation;

##ξ=-4x+6y## and ##η=6x+4y## where ##x=\dfrac{12ξ-18η}{156}## and ##y=\dfrac{3ξ+2η}{26}##

for the inhomogenous part, we shall have;

$$u_p = e^{\frac{-5}{26}ξ}\int \frac{e^{\frac{8ξ+η}{26}}⋅e^{\frac{5}{26}ξ}}{52} dξ$$$$u_p = \frac{e^{\frac{13ξ+η}{26}}⋅e^{\frac{-5}{26}ξ}}{26} $$

$$u(ξ,η)=u_h+u_p=e^{\frac{-5}{26} ξ} \left[ f(η)+\frac{e^{\frac{13ξ+η}{26}}}{26}\right]$$

$$u(x,y)=

e^{\frac{-5}{26} (-4x+6y)} \left[ f(6x+4y)+\frac{1}{26} {e^{\frac{-46x+82y}{26}}}\right]$$your insight is welcome or alternative approach to this problem.

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