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##### Chapter 2 / Impose Initial Condition on Inhomogeneous Equations

« Last post by**Yifei Hu**on

*February 01, 2022, 11:34:15 AM*»

Consider problem: $U_x+3U_y=xy, U(0,y)=0$, I proceed as follow:

$$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$$

Integrate on first two terms:

$$x-\frac{1}{3}y=C \qquad \color{red}{(*)}$$

Integrate on $\frac{du}{xy} = \frac{dx}{1}$: Error! You must remember that in the equation of characteristics $x$ and $y$ are not independent but connected by (*).

$$U =x^3-\frac{C}{2}x^2+D = \frac{3}{2}x^3+\frac{1}{2}yx^2+D $$

D must be constant along the integral curve hence $D=\phi(x-1/3y)$

Hence, the general solution is $U =\frac{3}{2}x^3+\frac{1}{2}yx^2+\phi(x-1/3y)$.

Impose initial condition: $U(0,y)=0$ we have $\phi(-1/3y)=0$.

Update:

Hi professor, I have fixed the integration part, but I still have question about the constant D when integrating on U. Should I include this $D=\phi(x-\frac{1}{3}y)$ in the solution to IVBP?

$$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$$

Integrate on first two terms:

$$x-\frac{1}{3}y=C \qquad \color{red}{(*)}$$

Integrate on $\frac{du}{xy} = \frac{dx}{1}$: Error! You must remember that in the equation of characteristics $x$ and $y$ are not independent but connected by (*).

$$U =x^3-\frac{C}{2}x^2+D = \frac{3}{2}x^3+\frac{1}{2}yx^2+D $$

D must be constant along the integral curve hence $D=\phi(x-1/3y)$

Hence, the general solution is $U =\frac{3}{2}x^3+\frac{1}{2}yx^2+\phi(x-1/3y)$.

Impose initial condition: $U(0,y)=0$ we have $\phi(-1/3y)=0$.

Update:

Hi professor, I have fixed the integration part, but I still have question about the constant D when integrating on U. Should I include this $D=\phi(x-\frac{1}{3}y)$ in the solution to IVBP?