Question

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

Answer 1

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.

Answer 2

\[\frac{\mathrm dy}{\mathrm dx}+3y=7\]
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
\[\frac{\mathrm d}{\mathrm dx}\big(\mu(x)y\big)=\mu(x) q(x)\]\[\mu(x)y=\int{\mu(x) q(x)\text dx} \]
\[y=\quad\dots\]

Answer 3

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

Answer 4

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

Answer 5

y = y(x)

the condition is y(0) = 0

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

Answer 6

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

Answer 7

or grammar apparently....

Answer 8

can you show some working

Answer 9

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

Answer 10

oh no wait a second....hold on

Answer 11

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

Answer 12

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

\[y(0)=0\]
\[0=\frac73+c\]\[c=-\frac73\]

\[y=\frac73-\frac73e^{-3x}\]

Answer 13

..............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!