Solve each linear 1st ordiary differential equation. Find explicit solutions y=f(x) if possible. If initial conditions are given, find the particular solution that satisfies them.
y' = y+2xe^2x y(0)=3

Question

Solve each linear 1st ordiary differential equation. Find explicit solutions y=f(x) if possible. If initial conditions are given, find the particular solution that satisfies them.
y' = y+2xe^2x        y(0)=3

anonymous · Answer

@satellite73 @jim_thompson5910

jim_thompson5910 · Answer

This is a good reference page I think http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx the first step is to get the equation into the form $$\Large \frac{dy}{dt} + p(t)*y = g(t)$$ tell me what you get when you do that step

anonymous · Answer

before doing that don't I have to state what P and Q is

anonymous · Answer

@jim_thompson5910

anonymous · Answer

P= -1 and Q=2xe^2x

jim_thompson5910 · Answer

I think you mean g(t) instead of q(t)

but yes,

p(t) = -1
g(t) = 2xe^(2x)

anonymous · Answer

well in my class we are using P and Q P=-1 and Q=2x2^2x

jim_thompson5910 · Answer

ok q works as well

jim_thompson5910 · Answer

Now you'll need to find the integrating factor `mu(t)`

$$\Huge \mu(t) = e^{{}^{\int p(t)dt}}$$

jim_thompson5910 · Answer

In your case, you're using x instead of t

anonymous · Answer

mu(x0=e^5ydx

anonymous · Answer

mu(x)*

jim_thompson5910 · Answer

$$\Huge \mu(x) = e^{{}^{\int p(x)dx}}$$

$$\Huge \mu(x) = e^{{}^{\int -1dx}}$$

$$\Huge \mu(x) = e^{-x}$$

anonymous · Answer

μ this sign is called mu right

jim_thompson5910 · Answer

yes that's the greek letter mu

anonymous · Answer

okay what i got for e above is tht right?

jim_thompson5910 · Answer

no the integrating factor should be e^(-x) and I show how I got that above

anonymous · Answer

oh okay so that's my N(x) yeah I see what I did wrong. also the greek letter mu could also be written as N(x) righ t

jim_thompson5910 · Answer

idk what you mean by N(x)

anonymous · Answer

nvm do not worry abotu that.

anonymous · Answer

okay so after finding mu I have to find general solution right which I do not know how to set up

jim_thompson5910 · Answer

$$\Large y \ ' = y + 2xe^{2x}$$

$$\Large y \ ' - y = 2xe^{2x}$$

$$\Large \color{red}{\mu(x)}*(y \ ' - y) = \color{red}{\mu(x)}*2xe^{2x}$$

$$\Large \color{red}{e^{-x}}*(y \ ' - y) = \color{red}{e^{-x}}*2xe^{2x}$$

$$\Large \frac{d}{dx}\left[e^{-x}*y\right] = e^{-x}*2xe^{2x}$$

$$\Large \frac{d}{dx}\left[e^{-x}*y\right] = 2xe^{-x+2x}$$

$$\Large \frac{d}{dx}\left[e^{-x}*y\right] = 2xe^{x}$$

$$\Large \int\frac{d}{dx}\left[e^{-x}*y\right]dx = \int 2xe^{x}dx$$

$$\Large e^{-x}*y  = \int 2xe^{x}dx$$


I'll let you finish up

anonymous · Answer

wow that's not even close to what I did.

anonymous · Answer

so the next step will be 2 integral xe^x dx

jim_thompson5910 · Answer

yes, use integration by parts

anonymous · Answer

parts?

jim_thompson5910 · Answer

integration by parts
$$\Large \int u dv = u*v - \int v du$$

jim_thompson5910 · Answer

u = x ---> du = dx
dv = e^x ---> v = e^x


$$\LARGE \int u dv = u*v - \int v du$$

$$\LARGE \int x e^x = x*e^x - \int e^x dx$$

$$\LARGE \int x e^x = x*e^x - e^x+C$$

anonymous · Answer

-x(e)(y)=2 integral xe
sorry I am getting confused this problem is supper long and the last one I have to deal with for now.

anonymous · Answer

u = x ---> du = dx
dv = e^x ---> v = e^x


∫udv=u∗v−∫vdu
this u showed is just an example right

jim_thompson5910 · Answer

yes that's just the basic template for integration by parts http://www.mathsisfun.com/calculus/integration-by-parts.html

anonymous · Answer

alright

anonymous · Answer

so integratign the left side will give me -e^-x y+c

anonymous · Answer

the right wil give us 2e^x (x-1)

jim_thompson5910 · Answer

I'd have the +C on the right side

I'm getting 

$$\Large e^{-x}*y  = 2x*e^x - 2e^x + C$$

jim_thompson5910 · Answer

2e^x (x-1) is equivalent to what I have on the right side (it's factored)

anonymous · Answer

according to your example it should be on the right only

anonymous · Answer

yeah I see it now

jim_thompson5910 · Answer

Multiply both sides by $\Large e^x$ to isolate y


$$\Large e^{-x}*y  = 2x*e^x - 2e^x + C$$

$$\Large e^{x}*e^{-x}*y  = e^{x}*\left(2x*e^x - 2e^x + C\right)$$

$$\Large y  = e^{x}*2x*e^x +e^{x}*(-2e^x) + e^{x}*C$$

$$\Large y  = 2x*e^{2x} - 2e^{2x} + Ce^{x}$$

$$\Large y(x)  = 2x*e^{2x} - 2e^{2x} + Ce^{x}$$

jim_thompson5910 · Answer

Now use y(0) = 3 to find the value of C

anonymous · Answer

what will be my general solution here?

jim_thompson5910 · Answer

I just wrote it

anonymous · Answer

yeah sorry I was reading it. okay so x=0 and y=3

jim_thompson5910 · Answer

the general solution will have a C in it
the particular solution will have C be replaced with some fixed number

anonymous · Answer

y(x)=2x∗e2x−2e2x+Cex
plugging that in your equation 
3(0)=2(0)(e^2(0)) - 2e^2(0)  + Ce^0

jim_thompson5910 · Answer

yes, then simplify and solve for C

anonymous · Answer

0=0(1)-2+c
1=c?

jim_thompson5910 · Answer

nope

jim_thompson5910 · Answer

y(0) = 3, so the LHS should be 3

jim_thompson5910 · Answer

3 = -2 + C

anonymous · Answer

ph okay so c=5 what is LHS angain?

jim_thompson5910 · Answer

LHS = left hand side

jim_thompson5910 · Answer

yes C = 5

anonymous · Answer

alright tysm!

jim_thompson5910 · Answer

you're welcome