MHB 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:

$$\frac{dy}{dx}-\frac{1}{4}y=\frac{1}{4}e^{\frac{x}{4}}$$

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:

$$\frac{dy}{dx}-\frac{1}{4}y=\frac{1}{4}e^{\frac{x}{4}}$$

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 $$P(x)=-\frac{1}{4}$$.

 

[alyafey22]

$$4y' - y = e^{\frac{x}{4}}$$

Divide by $$4 e^{\frac{x}{4}}$$

$$y'e^{-\frac{x}{4}} - \frac{1}{4}\,y\,e^{-\frac{x}{4}} = \frac{1}{4}$$

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

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

Integrating with respect to x :

$$y\, e^{-\frac{x}{4}}=\frac{1}{4}x\,+\,C$$

$$y\,=\frac{1}{4}\,x\, e^{\frac{x}{4}}+\,C e^{\frac{x}{4}}$$

 

[Logan Land]

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

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
\frac{dY}{dx}= f(x)

Initially we have
4\frac{dy}{dx}- y= e^{x/4}
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
\frac{dy}{dx}- y/4= e^{x/4}/4

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

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

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

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