-206.q3.2 method of integrating factor

[karush]

2000
$\tiny{206.q3.2}\\$
$\textsf{3. use the method of integrating factor}\\$
$\textsf{to find the general solution to the first order linear differential equation}\\$
\begin{align}
\displaystyle
\frac{dy}{dx}+5y=10x
\end{align}
$\textit{clueless !}$

 

[MarkFL]

Given a first order linear ODE of the form:

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

We can use an integrating factor $\mu(x)$ to make the LHS of the ODE into the derivative of a product, using the special properties of the exponential function with regard to differentiation. Consider what happens if we multiply though by:

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

We get:

\(\displaystyle \exp\left(\int f(x)\,dx\right)\d{y}{x}+\exp\left(\int f(x)\,dx\right)f(x)y=g(x)\exp\left(\int f(x)\,dx\right)\)

Now, let's use:

\(\displaystyle F(x)=\int f(x)\,dx\implies F'(x)=f(x)\)

And we now have:

\(\displaystyle \exp\left(F(x)\right)\d{y}{x}+\exp\left(F(x)\right)F'(x)y=g(x)\exp\left(F(x)\right)\)

Now, if we observe that, via the product rule, we have:

\(\displaystyle \frac{d}{dx}\left(\exp(F(x))y\right)=\exp\left(F(x)\right)\d{y}{x}+\exp\left(F(x)\right)F'(x)y\)

Then, we may now write our ODE as:

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

Now, we may integrate both sides w.r.t $x$.

So, in the given ODE:

\(\displaystyle \d{y}{x}+5y=10x\)

We identify: \(\displaystyle f(x)=5\)

And so we compute the integrating factor as:

\(\displaystyle \mu(x)=\exp\left(5\int\,dx\right)=\)?

 

[karush]

wow that was a great help. but I'll have go thru this tonmorro ...

ok, got it, but have never come up with that