Solving a linear first order differential equation

[Logan Land]

4y'=e^(x/4) + y

First I need to divide through by 4 correct?
To obtain
y'=(e^(x/4))/4 + (y/4)

But then when I try to find integrating factor I just come up with e^(x/4) which I think is incorrect

 

[alyafey22]

Re: Solving a differential equation

First put the equation in the standard form.

 

[MarkFL]

I have edited the title to make it a bit more descriptive of the problem. A topic's title should give at least one level of information more than is implied by the forum in which it is posted. :D

I would first arrange the ODE in standard linear form by subtracting through by $y$ and then dividing through by 4:

\(\displaystyle \frac{dy}{dx}-\frac{1}{4}y=\frac{1}{4}e^{\frac{x}{4}}\)

Now, what is your integrating factor $\mu(x)$?

 

[Logan Land]

I have edited the title to make it a bit more descriptive of the problem. A topic's title should give at least one level of information more than is implied by the forum in which it is posted. :D

I would first arrane the ODE in standard linear form by subtracting through by $y$ and then dividing through by 4:

\(\displaystyle \frac{dy}{dx}-\frac{1}{4}y=\frac{1}{4}e^{\frac{x}{4}}\)

Now, what is your integrating factor $\mu(x)$?

I still keep getting e^(x/4)

Do I not put it as e^(1/4)(integral e^(x/4))

 

[alyafey22]

Don't forget the sign !.

 

[MarkFL]

Don't forget the sign !.

Yes, in this problem, we have \(\displaystyle P(x)=-\frac{1}{4}\).

 

[alyafey22]

\(\displaystyle 4y' - y = e^{\frac{x}{4}}\)

Divide by \(\displaystyle 4 e^{\frac{x}{4}}\)

\(\displaystyle y'e^{-\frac{x}{4}} - \frac{1}{4}\,y\,e^{-\frac{x}{4}} = \frac{1}{4}\)

Now we realize that \(\displaystyle \frac{d}{dx} \left( y e^{-\frac{x}{4}}\right) = y'e^{-\frac{x}{4}} -\frac{1}{4} y\,e^{-\frac{x}{4}} \)

Hence \(\displaystyle \frac{d}{dx} \left( y e^{-\frac{x}{4}}\right)=\frac{1}{4}\)

Integrating with respect to x :

\(\displaystyle y\, e^{-\frac{x}{4}}=\frac{1}{4}x\,+\,C\)

\(\displaystyle y\,=\frac{1}{4}\,x\, e^{\frac{x}{4}}+\,C e^{\frac{x}{4}}\)

 

[Logan Land]

Yes, in this problem, we have \(\displaystyle P(x)=-\frac{1}{4}\).

So no matter what we use the information on the left to compute the integrating factor? Even in this case with no x just (-1/4)

In that case would it be (-1/4)x?

e^(-1/4)integral dxOh the integrating factor is e^(-x/4)?

 

[alyafey22]

Oh the integrating factor is e^(-x/4)?

yes .

 

[HallsofIvy]

You want something of the form
[tex]\frac{dY}{dx}= f(x)[/tex]

Initially we have
[tex]4\frac{dy}{dx}- y= e^{x/4}[/tex]
where I have put all the "y" dependence on the left.

Yes, you can if you like divide by 4- that is not necessary but makes the calculations simpler
[tex]\frac{dy}{dx}- y/4= e^{x/4}/4[/tex]

We want to multiply by some u(x) that will make the left side a single derivative: that the left side becomes [tex]\frac{d(uy)}{dx}[/tex].

By the product rule, [tex]\frac{d(uy)}{dx}= u\frac{dy}{dx}+ \frac{du}{dx}y[/tex] and, we can see that, by multiplying by u we would have [tex]u\frac{dy}{dx}- \frac{uy}{4}[/tex]. Comparing those, we see we must have [tex]\frac{du}{dx}= -\frac{u}{4}[/tex], a simple, separable equation: [tex]\frac{du}{u}= -\frac{dx}{4}[/tex] and integerating [tex]ln(u)= -\frac{x}{4}+ C[/tex] or [tex]u= Ce^{-x/4}[/tex].

That is, an "integrating factor" is [tex]e^{-x/4}[/tex]:
[tex]e^{-x/4}\frac{dy}{dx}- \frac{e^{-x/4}y}{4}= \frac{d(e^{-x/4}y}{dx}[/tex]

so multiplying both sides of the original equation by [tex]e^{-x/4}[/tex] we have
[tex]\frac{d(e^{-x/4}y}{dx}= \frac{1}{4}[/tex]
and we can integrate both sides to get
[tex]e^{-x/4}y= \frac{x}{4}+ C[/tex]
so that [tex]y= \frac{x}{4}e^{x/4}+ Ce^{x/4}[/tex].