Differential Equations Examples

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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 $y=3e^{2t}.$

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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 $y=1+{\frac {C}{e^{t}}}$.

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