Pessimist Singularitarian
#2
December 10th, 2016,
03:07
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)=$?