Here's the question you clicked on:
chambek
dy/dx+3y=7 y(0)=0 Find a particular solution to the nonhomogeneous differential equation.
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.
\[\frac{\mathrm dy}{\mathrm dx}+3y=7\] 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 \[\frac{\mathrm d}{\mathrm dx}\big(\mu(x)y\big)=\mu(x) q(x)\]\[\mu(x)y=\int{\mu(x) q(x)\text dx} \] \[y=\quad\dots\]
after integration you will have a constant of integration, substituting the initial value will allow you to find it
I got the first part! Thanks, I'm trying to get the part with the initial value now....where do i substitute that in?
y = y(x) the condition is y(0) = 0 which means (x,y) = (0,0) is a solution
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
or grammar apparently....
can you show some working
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
oh no wait a second....hold on
grrr nevermind, that still got me 0, I had made a mistake in my algebra......
\[\frac{\mathrm dy}{\mathrm dx}+3y=7\]\[\mu(x)=e^{\int 3\text dx}=e^{3\int \text dx}=e^{3x}\]\[\frac{\mathrm d}{\mathrm dx}\big(e^{3x}y\big)=7e^{3x}\] \[e^{3x}y=\int{7 e^{3x}\text dx}=7\int e^{3x}\text dx=7\frac{e^{3x}}3+c\] \[y=\frac73+ce^{-3x}\] \[y(0)=0\] \[0=\frac73+c\]\[c=-\frac73\] \[y=\frac73-\frac73e^{-3x}\]
..............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!