Wize University Linear Algebra Textbook > Differential Equations

Basics of Differential Equations

Linear Dynamical SystemsPrevious Section

Systems of Linear Differential EquationsNext Section

0:00 / 0:00

Definition of a Linear Differential Equation
A differential equation is an equation relating a function and its derivatives. The goal is to solve for this unknown function!

The order of a differential equation is the highest derivative it contains.

Examples
y′=3yy'=3yy′=3y is a first order differential equation
y′′+5y′−3y=0y''+5y'-3y=0y′′+5y′−3y=0 is a second order differential equation
These are also examples of linear differential equations, since yyy (and its derivatives) are only ever multiplied by constants.

Note: Differential equations can be non-linear, for example:

(y′′)2−ln⁡(y′)=3y(y'')^2-\ln(y')=3y(y′′)2−ln(y′)=3y

We will only be focusing on linear differential equations.

0:00 / 0:00

Simplest Differential Equation: y′=ay\colorOne{y'=ay}y′=ay
The simplest differential equation (DE) is:
y′=ayy'=ayy′=ay
You should memorize the solution to this equation:
y′=ay  ⟹  y(x)=Ceaxy'=ay \qquad \implies \qquad
\boxed{\quad
y(x)=Ce^{ax}
\quad}y′=ay⟹y(x)=Ceax​
where CCC is an arbitrary constant (it can be any number).
Wize Tip
Note that the solution to a differential equation is always a function.

Check
To verify the solution, we plug it into the DE and check that LHS=RHS.

y=Ceax  ⟹  LHS of DERHS of DEy′=  a(Ceax)ay=  a(Ceax)y = Ce^{ax} \quad \implies \quad 
\begin{array}{c |c}
\text{LHS of DE} \quad &\quad \text{RHS of DE}\\
\hline\\[-1em]
\begin{aligned}
&y' \\
=\ \ &a(Ce^{ax})
\end{aligned}

&

\begin{aligned}
&ay \\
=\ \ &a(Ce^{ax})
\end{aligned}

\end{array}y=Ceax⟹LHS of DE=  ​y′a(Ceax)​​RHS of DE=  ​aya(Ceax)​​​
Since LHS=RHS, we've shown that our solution does indeed satisfy y′=ayy'=ayy′=ay.

PAGE BREAK
Examples
a) y′=−7yy'=-7yy′=−7y. What function satisfies this relationship?

y=Ce−7xy=Ce^{-7x}y=Ce−7x

Check your answer by taking the derivative of your solution.

y=Ce−7x  ⟹  y′=−7(Ce−7x)⏟y=−7yy=Ce^{-7x} \quad \implies \quad
y' = -7\underbrace{\colorOne{(Ce^{-7x})}}_{\normalsize y}=-7yy=Ce−7x⟹y′=−7y(Ce−7x)​​=−7y

b) y′=35yy'=\frac{3}{5}yy′=53​y. What is the solution?

y=Ce35xy=Ce^{{\normalsize \frac{3}{5}}x}y=Ce53​x

0:00 / 0:00

Initial Value Problems
We saw that the solution to the differential equation y′=ayy'=ayy′=ay  is  y=Ceaxy=Ce^{ax}y=Ceax.
If the problem provides initial conditions (an initial value for yyy), then we can find a specific value for CCC.
A differential equation with an initial condition is called an initial value problem.
Example
Find the function yyy such that y′=5yy'=5yy′=5y and y(0)=4y(0)=4y(0)=4.
Start by writing the general solution to this DE:

y=Ce5xy=Ce^{5x}y=Ce5x

Since we want y(0)=4y(0) = 4y(0)=4, we know that y=4y=4y=4 when x=0x=0x=0. 
Plug these into our solution, and solve for CCC:

4=Ce5(0)4=Ce04=C(1)  ⟹  4=C\begin{aligned}
4&=Ce^{5(0)}\\
4&=Ce^0\\
4&=C(1) \\
\implies 4&=C
\end{aligned}444⟹4​=Ce5(0)=Ce0=C(1)=C​
Therefore, the solution is y=4e5x\boxed{y=4e^{5x}}y=4e5x​

0:00 / 0:00

Example: Solving an Initial Value Problem
Find the function such that y′=−6yy'=-6yy′=−6y  with  y(0)=ln⁡(2)y(0)=\ln(2)y(0)=ln(2).

y′=−6y→has solutiony=Ce−6xy'=-6y \quad \xrightarrow{\text{has solution}} \quad
 y=Ce^{-6x}y′=−6yhas solution​y=Ce−6x
We want y=ln⁡(2)y=\ln(2)y=ln(2) when x=0x=0x=0. Plugging these values into our solution, we solve for CCC:
ln⁡(2)=Ce−6(0)ln⁡(2)=Ce0  ⟹  ln⁡(2)=C\begin{aligned}
\ln(2)&=Ce^{-6(0)}\\
 \ln(2)&=Ce^0\\
\implies \ln(2)&=C
\end{aligned}ln(2)ln(2)⟹ln(2)​=Ce−6(0)=Ce0=C​
So the solution is therefore: y=ln⁡(2)e−6x\boxed{y=\ln(2)e^{-6x}}y=ln(2)e−6x​

Practice: Simple Differential Equations
For each differential equation (DE), find the general solution (i.e. the function that satisfies it):

a) y1′=−3y1y_1'=-3y_1y1′​=−3y1​

b) y2′=−12y2y_2'=-\frac{1}{2}y_2y2′​=−21​y2​

c) y3′=5y3y_3'=5y_3y3′​=5y3​

(Don't forget the arbitrary constant, CCC)

y_1(x)=

y_2(x)=

y_3(x)=

I don't know

Practice: Initial Value Problems
Find the function that satisfies the differential equation y′=12yy'=\frac{1}{2}yy′=21​y  with initial condition  y(2)=1y(2)=1y(2)=1.

A) y=e(12x−1)y= e^{\left({\normalsize\frac{1}{2}}x - 1 \right)}y=e(21​x−1)

B) y=12e12xy= \dfrac{1}
{2}e^{{\normalsize\frac{1}{2}}x}y=21​e21​x

C) y=Ce12xy= Ce^{{\normalsize\frac{1}{2}}x}y=Ce21​x

D) y=2e12xy= 2e^{{\normalsize\frac{1}{2}}x}y=2e21​x

I don't know

Extra Practice

Practice: Linear DE

Q:\bf{Q:}Q: Solve  x2  ⁣dy  ⁣dx+3xy=1x^2 \dfrac{\de{y}}{\de{x}} + 3xy = 1x2 dx dy​+3xy=1 
  

Find a solution to the differential equation 
dydt=3y\frac{dy}{dt}=3ydtdy​=3y
where y(0)=6y(0) = 6y(0)=6 .

Solve the differential equation 
dydx+(tan⁡ x)y =sin⁡x \frac{dy}{dx}+\left(\tan\ x\right)y\ =\sin x\ dxdy​+(tan x)y =sinx 
subject to y(2π)=π3.y\left(2\pi\right)=\frac{\pi}{3}.y(2π)=3π​.

Differential Equations

The functions  y=x2+cx2y=x^2+\frac{c}{x^2}y=x2+x2c​, are all solutions of equation: xy′+2y=4x2, x>0.xy'+2y=4x^2,\ x>0.xy′+2y=4x2, x>0.
Find the constant ccc  which produces a solution which also satisfies the initial condition  y(6)=5.y\left(6\right)=5.y(6)=5.

Differential Equations

Find an equation of the curve that satisfies  dydx=80yx15\frac{dy}{dx}=80yx^{15}dxdy​=80yx15and whose y-intercept is 666.

Differential Equations

Let y′=7xy'=7xy′=7x. Find all values of rrr such that y=rx2y=rx^2y=rx2 satisfies the differential equation.
List all the values separated by a comma.

Differential Equations

Solve the differential equationdydx=4yx\frac{dy}{dx}=\frac{4y}{x}dxdy​=x4y​,x>0.x>0.x>0.

Wize University Linear Algebra Textbook > Differential Equations

Basics of Differential Equations

Popular Courses

Definition of a Linear Differential Equation

Simplest Differential Equation: $\colorOne{y'=ay}$

Initial Value Problems

Example: Solving an Initial Value Problem

Practice: Simple Differential Equations

Practice: Initial Value Problems

Extra Practice

Practice: Linear DE

Differential Equations

Differential Equations

Differential Equations

Differential Equations