-2.1.1 solve y'+3y = x + e^{-2x}

[karush]

\nmh{1000}
$$\displaystyle y^\prime + 3y = x + e^{-2x}$$
$\textit{Solve the given differential equation}$
\begin{align*}\displaystyle
&\textit{book answer is}\\
y&=\color{red}{\disnmplaystyle ce^{-3x}+\frac{x}{3}-\frac{1}{9}+e^{-2x}}
\end{align*}
no sure how $ce^{-3x}$ was derived?

 

[MarkFL]

Typically, when given a linear first order ODE of the form:

\(\displaystyle \d{y}{x}+f(x)y=g(x)\)

You want to look for an integrating factor in the form:

\(\displaystyle \mu(x)=\exp\left(\int f(x)\,dx\right)\)

What would your integrating factor be for this problem?

 

[karush]

Typically, when given a linear first order ODE of the form:

\(\displaystyle \d{y}{x}+f(x)y=g(x)\)

You want to look for an integrating factor in the form:

\(\displaystyle \mu(x)=\exp\left(\int f(x)\,dx\right)\)

What would your integrating factor be for this problem?

$\d{y}{x}+Py=Q$not real sure but if P=3 then
$$I=e^{\int 3 \, dx}$$

 

[MarkFL]

$\d{y}{x}+Py=Q$not real sure but if P=3 then
$$I=e^{\int 3 \, dx}$$

Simplify it by evaluating the integral (without the constant of integration). What do you get?

 

[MarkFL]

With the integrating factor defined as:

\(\displaystyle \mu(x)=\exp\left(\int f(x)\,dx\right)\)

It then follows that:

\(\displaystyle \mu'(x)=\exp\left(\int f(x)\,dx\right)f(x)\)

And thus, when you multiply the general ODE I posted by this factor, you obtain:

\(\displaystyle \mu(x)\d{y}{x}+\mu'(x)y=\mu(x)g(x)\)

We should observe now that the LHS can be rewritten as the derivative of a product:

\(\displaystyle \frac{d}{dx}\left(\mu(x)y(x)\right)=\mu(x)g(x)\)

And now it is possible to solve for $y(x)$ after integrating. I will await your attempt to apply this method to the given problem. :)

 

[karush]

With the integrating factor defined as:

\(\displaystyle \mu(x)=\exp\left(\int f(x)\,dx\right)\)

It then follows that:

\(\displaystyle \mu'(x)=\exp\left(\int f(x)\,dx\right)f(x)\)

And thus, when you multiply the general ODE I posted by this factor, you obtain:

\(\displaystyle \mu(x)\d{y}{x}+\mu'(x)y=\mu(x)g(x)\)

We should observe now that the LHS can be rewritten as the derivative of a product:

\(\displaystyle \frac{d}{dx}\left(\mu(x)y(x)\right)=\mu(x)g(x)\)

And now it is possible to solve for $y(x)$ after integrating. I will await your attempt to apply this method to the given problem. :)

i presume you imply taking the integral of both sides first
$$\frac{d}{dx}\left(\mu(x)y(x)\right)=\mu(x)g(x)$$
$$\mu(x)y(x)=\int\mu(x)g(x) \, dx $$

 

[MarkFL]

i presume you imply taking the integral of both sides first
$$\frac{d}{dx}\left(\mu(x)y(x)\right)=\mu(x)g(x)$$
$$\mu(x)y(x)=\int\mu(x)g(x) \, dx $$

Yes, that's what you have after integrating. And then you may state:

\(\displaystyle y(x)=\frac{1}{\mu(x)}\left(\int \mu(x)g(x)\,dx\right)\)

Can you apply this technique to the given problem?

 

[karush]

Yes, that's what you have after integrating. And then you may state:

\(\displaystyle y(x)=\frac{1}{\mu(x)}\left(\int \mu(x)g(x)\,dx\right)\)

Can you apply this technique to the given problem?

ok I think I'm lost on $\mu(x)g(x)$
\begin{align}\displaystyle
\frac{d}{dx}\left(\mu(x)y(x)\right)&=\mu(x)g(x)\\
y^\prime &= x + e^{-2x}-3y\\
y+3&=\frac{1}{x}+e^{-2x}\\
book \, answer&\\
y&=\color{red}{\displaystyle ce^{-3x}+\frac{x}{3}-\frac{1}{9}+e^{-2x}}
\end{align}

 

[MarkFL]

We have the integrating factor:

\(\displaystyle \mu(x)=e^{3x}\)

Multiplying the given ODE by this factor we obtain:

\(\displaystyle e^{3x}y'+3e^{3x}y=xe^{3x}+e^{x}\)

Rewriting the LHS, we now have:

\(\displaystyle \frac{d}{dx}\left(e^{3x}y\right)=xe^{3x}+e^{x}\)

Integrating w.r.t $x$, we get:

\(\displaystyle e^{3x}y=\frac{1}{9}e^{3x}(3x-1)+e^x+c_1\)

Hence:

\(\displaystyle y(x)=\frac{1}{9}(3x-1)+e^{-2x}+c_1e^{-3x}\)

This is equivalent to the answer given by your book. :)

When you study 2nd (and higher) order linear ODE's, you will be shown other methods to solve problems like this.

 

[karush]

wow, thanks for the awesome help
I don't start a class in this for 10 weeks yet
but just want to get a head start on it.

this is the best place to come for that.

 

[DrWahoo]

Using an integrating factor helps make the differential equations "Exact". Wonderful way to solve simple and some decently complex ode's if you can integrate the integrating factor. Can you check that once you multiply by the integrating factor that you do indeed have an exact ode?