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First Order Linear Differential Equations

April 5th, 2009 |

Author: Merrin

First order linear differential equations are another class of differential equations which potentially have analytic solutions. All such equations can be written in the form

$\displaystyle \cfrac{dy}{dx} +p(x)y=q(x)$

When q(x)=0 the equation is said to be homogeneous and when it’s not the equation is said to be nonhomogeneous.

One method to solve this equation is through an integrating factor. An integration factor multiplies both sides of the equation such that terms group together as if they were a perfect derivative. Here is what that means.

$\displaystyle I(x)\cfrac{dy}{dx} +I(x)p(x)y(x)=\cfrac{d}{dx} (I(x)y(x))=I(x)q(x)$

If we could find the integrating factor, I(x), we could just integrate both sides with the result that

$\displaystyle I(x)y(x)=\int_b^x I(t)q(t)dt +C$
$\displaystyle y=\cfrac{C}{I(x)} +\cfrac{1}{I(x)} \int_b^x I(t)q(t)dt$

Now let us focus in on what this integrating factor might be.

$\displaystyle I\cfrac{dy}{dx} +Ip(x)y=\cfrac{d}{dx} (Iy)$

One guess might be

$I(x)=e^{p(x)}$
$I(x)'=p'e^{p} =p'I$

Which gives,

$\cfrac{d}{dx} (Iy)=Iy'+Ip'y$

We were so close. By close I mean compared to our target of

$Iy'+Ipy$

We need to get rid of that derivative and replace it with $p(x)$. It is kind of tricky how we do it. Try

$\displaystyle I(x)=\exp (\int_{a}^{x}p(t)dt )=e^{\int_{a}^{x}p(t)dt }$

Recall Leibniz’s rule for differentiating integrals with variable limits.

$\displaystyle \cfrac{d}{dx} \int _{v(x)}^{u(x)}f(t,x)dt =u'f(u,x)-v'f(v,x)+\int _{v(x)}^{u(x)}\cfrac{\partial }{\partial x} f(t,x)dt$

Luckily our expression is not so complicated that we have to use all of this formula. We can calculate the derivative of I(x).

$\displaystyle I'(x)=I(x)\cfrac{d}{dx} \int _{b}^{x}p(t)dt=I(x)p(x)$

What is b? This is just any constant you choose which leads to an overall constant multiplying the integration factor. It doesn’t change the form of the solution.

Now we are set. We have

$\cfrac{d}{dx} (Iy)=I'y+Iy'=I(y'+py)$

We have already put the general solution previously and now we have an equation for I(x) so we are done.

$\displaystyle I(x)=\exp (\int _{a}^{x}p(t)dt )=e^{\int _{a}^{x}p(t)dt }$

Perhaps the most concise representation of the integration factor is to write it as

$I(x)=e^P$

where capital P is an antiderivative of lowercase p.

$\displaystyle y=\cfrac{C}{I(x)} +\cfrac{1}{I(x)} \int_b^x I(t)q(t)dt$

Example 1. Find the solution to the following differential equation

$\cfrac{dy}{dx} +x^{2} y=x^{2}$

Solution 1. We form the integrating factor as

$\displaystyle {I(x)=e^{\int _{a}^{x}t^{2} dt } =c_{1} e^{x^{3} /3} \to e^{x^{3} /3} \quad \, Let \,\,c_{1} =1}$
${\cfrac{d}{dx} (yI)=Ix^{2} }$
$\displaystyle y(x)I(x)=\int_b^x e^{t^{3}/3} t^2 dt+C=\int_{b^*}^{x^3/3} e^{u} du+C=e^{x^{3}/3}+C$
$y(x)=1+Ce^{-x^{3} /3}$

It is always possible to substitute the solution back into the differential equation and indeed verify that it is a solution.

Example 2. Find the solution to the following differential equation
$\cfrac{dy}{dx} +(\ln x)y=e^{x} (\ln x+1)$

Solution 2. Wow, look at all those natural logarithms and exponentials, but don’t give up hope. The answer will come together in the end. We use the integrating factor to find the solution

$\displaystyle {I(x)=\exp [\int _{b}^{x}\ln tdx ]=\exp [(x\ln x-x)-(b\ln b-b)]}$
${b=e}$
${I(x)=x^x e^{-x} }$
$\displaystyle {y=\cfrac{C_1}{x^{x} e^{-x} } +\cfrac{1}{x^{x} e^{-x} } \int x^{x} e^{-x} (e^{x} (\ln x+1))dx }$
$\displaystyle {y=C_1x^{-x} e^{x} +\cfrac{e^{x} }{x^{x} } \int x^{x} (\ln x+1)dx }$

Recall the derivative of x to the x is just

$x^x(\ln x +1)$
$\cfrac{e^x}{x^x} \int x^x(\ln x+1)dx =\cfrac{e^{x} }{x^x}(x^x+C_2)$

The two integration constants can just be combined into one for the general solution.

$\displaystyle y=Cx^{-x} e^{x} +e^{x}=e^x\left(1+C/x^x\right)$

The road to a closed form solution for first order linear differential equations usually goes through the path of

$\displaystyle \int_a^x p(s)ds \quad \text{and} \quad \int_b^x I(t)q(t)dt$

If it is not known how to solve these integrals it is not a complete failure because the answer in the closed form of definite integrals.

$y(x) = \cfrac{C}{e^{\int_a^xp(s)ds}}+\cfrac{1}{e^{\int_a^xp(s)ds)}}\int_b^x e^{\int_a^tp(s)ds} q(t)dt$

Now we have solved another class of differential equations.

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One Response to “First Order Linear Differential Equations”

Ted Burrett says:

April 24, 2009 at 1:26 pm

If you want to read a reader’s feedback , I rate this article for four from five. Decent info, but I have to go to that damn msn to find the missed bits. Thank you, anyway!

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First Order Linear Differential Equations

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