solving first order ODE with integrating factor

Posted: April 3, 2013 in Mathematics

Personally, I don’t like DEs because they are boring, a bit too much of calculation for my liking.  Maybe this is my first and last post on DEs here.

Anyway, this is for a friend’s request…

So how do we solve non-separable ODE of this type?

\dfrac{dy}{dx}+p(x)y=q(x)

The answer is multiplying both sides by the integrating factor.

e^{\int p(x)\, dx}

The idea comes from the product rule and this, since

\frac{d}{dx}ye^x = y'e^x+ye^x

and if we replace the exponent of e with \int p(x)\,dx, then

\dfrac{d}{dx}e^{\int p(x)\, dx}=p(x)e^{\int p(x)\, dx} by chain rule

Therefore

y'+p(x)y=q(x)

becomes

y'e^{\int p(x)\, dx}+p(x)ye^{\int p(x)\, dx}=q(x)e^{\int p(x)\, dx}

\dfrac{d}{dx} ye^{\int p(x)\, dx} = q(x)e^{\int p(x)\, dx}

then

\displaystyle y=e^{-\int p(x)\, dx}\int q(x)e^{\int p(x)\, dx} \,dx

For example

y'+y=e^{-x}, \quad y(0)=1

then the integrating factor is

e^{\int p(x)\, dx} = e^{\int dx} = e^x

hence, we have

y'e^x+ye^x=e^{-x}e^x

\dfrac{d}{dx} ye^x = 1

\displaystyle ye^x=\int dx = x+C

since the initial value y(0)=1, we have

(1)e^0=0+C

C=1

Therefore

ye^x=x+1

y=\dfrac{x+1}{e^x}

 

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