## chambek 2 years ago dy/dx+3y=7 y(0)=0 Find a particular solution to the nonhomogeneous differential equation.

1. chambek

Also.....Find the solution to the initial value problem. I'm trying to learn this on my own but really am just not getting it.

2. UnkleRhaukus

$\frac{\mathrm dy}{\mathrm dx}+3y=7$ is a linear equation of the from $\frac{\mathrm dy}{\mathrm dx}+p(x)y=q(x)$ to solve the linear equation we need an integrating factor $\mu(x)=e^{\int p(x)\text dx}$ then $\frac{\mathrm d}{\mathrm dx}\big(\mu(x)y\big)=\mu(x) q(x)$$\mu(x)y=\int{\mu(x) q(x)\text dx}$ $y=\quad\dots$

3. UnkleRhaukus

after integration you will have a constant of integration, substituting the initial value will allow you to find it

4. chambek

I got the first part! Thanks, I'm trying to get the part with the initial value now....where do i substitute that in?

5. UnkleRhaukus

y = y(x) the condition is y(0) = 0 which means (x,y) = (0,0) is a solution

6. chambek

ok....doing that I got 0....and that was wrong, it is asking for a y=..... I'm sorry i dont understand my professor at all so i essentially now nothing about diffe q's

7. chambek

or grammar apparently....

8. UnkleRhaukus

can you show some working

9. chambek

sure, so i got (7/3) as the particular solution to the nonhomogeneous equation. then going from there, I included a C in my integral so I could plug in 0 for y and solce for c, then get an equation for y=.... $3y=7+C$ I found C to be -7 at y(0)=0 giving me $3y=7-7$ which just doesnt really make too much sense, but I tried 0 regardless and realized I have no clue what I'm doing haha

10. chambek

oh no wait a second....hold on

11. chambek

grrr nevermind, that still got me 0, I had made a mistake in my algebra......

12. UnkleRhaukus

$\frac{\mathrm dy}{\mathrm dx}+3y=7$$\mu(x)=e^{\int 3\text dx}=e^{3\int \text dx}=e^{3x}$$\frac{\mathrm d}{\mathrm dx}\big(e^{3x}y\big)=7e^{3x}$ $e^{3x}y=\int{7 e^{3x}\text dx}=7\int e^{3x}\text dx=7\frac{e^{3x}}3+c$ $y=\frac73+ce^{-3x}$ $y(0)=0$ $0=\frac73+c$$c=-\frac73$ $y=\frac73-\frac73e^{-3x}$

13. chambek

..............so, i got that awhile ago, and I kept putting the answer in as you have it, but, the equation is in terms of t.......*facepalm* thanks for the help!