# Differential Equations Examples This article is part of the MathHelp Tutoring Wiki

## Initial Value Problem

Example 1 :
Question:

Solve the following initial value problem.
$y'=3t^{2}-4$ , y(0) = 5
Given that the general solution is $f(t)=t^{3}-4t+C$ Solution:
Since we know y(0) = 5.
That means t = 0.
Therefore $f(0)=0^{3}-4(0)+C=5$ $0-0+C=5$ $C=5$ The desired solution is $f(t)=t^{3}-4t+5.$ Example 2 :
Question:

Solve the following initial value problem.
$y'=2y$ , y(0) = 3
Given that the general solution is $y=Ce^{2t}$ Solution:
Since we know y(0) = 3.
That means t = 0.
Therefore $y(0)=Ce^{2(0)}=3$ $C(1)=3$ $C=3$ The desired solution is $y=3e^{2t}.$ ## First-Order Linear Differential Equations

To solve first-order differential equations, you'll need to know the following facts:
• A first-order linear differential equation in standard form is y' + a(t)y = b(t).
• Find an integrating factor by integrating e^(∫a(t)dt) which will become the integrating factor in the form $e^{A(t)}$ .
• After find the integrating factor, multiply the first-order linear differential equation by the integrating factor.
• Then be able to recognize how to combine terms in y' and y into a single term after multiplying the differential equation by the integrating factor, for example:
1. $e^{t}y'+e^{t}y={\frac {d}{dt}}(e^{t}y)$ 2. $t^{2}y'+2ty={\frac {d}{dt}}(t^{2}y)$ • After combining the terms into a single term, integrate both sides to solve for y.
• By solving y, you can find the general solution for y.

Example 1 :
Question:

Solve $t^{2}y'+ty$ .
Solution:
Divide both sides by $t^{2}$ to ensure the differential equation is in its standard form (y' + a(t)y = b(t)).
So the differential equation turns into $y'+{\frac {1}{t}}y={\frac {2}{t^{2}}}$ Find the integrating factor by integrating e^(∫a(t)dt) = e^(∫${\frac {1}{t}}dt)=e^{ln(t)}$ Then the integrating factor is $e^{A(t)}=e^{ln(t)}=t$ .
Multiply the differential equation by the integrating factor t.
$(t)y'+(t){\frac {1}{t}}y=(t){\frac {2}{t^{2}}}=ty'+y={\frac {2}{t}}$ Recognize that the left hand side has the terms y' and y and you can combine them into a single term.
After combining it into a single term, it is ${\frac {d}{dt}}(ty)={\frac {2}{t}}$ .
You'll probably notice that you can combine the terms by recognizing that the single term is y times integrating factor = $e^{A(t)}$ and in this example, it is yt.
Now integrate both sides and you'll get ty = ∫${\frac {2}{t}}dt=2lnt+C$ .
Now isolate y and get $y={\frac {2lnt+C}{t}}$ .
The general solution is $y={\frac {2lnt+C}{t}}$ .

Example 2 :
Question:

Solve $y'+y=1$ .
Solution:
We know that a(t) = 1 and b(t) = 1.
Integrating factor = e^(∫1dt) = $e^{t}$ .
$e^{t}y'+e^{t}y=e^{t}$ ${\frac {d}{dt}}(e^{t}y)=e^{t}$ $e^{t}y=$ $e^{t}dt+C$ $e^{t}y=e^{t}+C$ $y={\frac {e^{t}+C}{e^{t}}}$ $y=1+{\frac {C}{e^{t}}}$ $y=1+e^{-t}C$ The general solution is $y=1+e^{-t}C$ or $y=1+{\frac {C}{e^{t}}}$ .