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Wize University Linear Algebra Textbook > Differential Equations

Higher Order Differential Equations

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Higher Order Linear DEs

A second order differential equation (DE) involves a function along with its first and second derivatives.
E.g. y′′+3y′−y=0y''+3y'-y=0
Such equations can be converted into a system of first order DEs by making the following substitutions:
→\rightarrowLet y1=yy_1=y and y2=y′y_2=y'
Note: Given a third order differential equation, make the substitution y3=y′′y_3=y'' (and so on for higher orders).
With these substitutions, we can write an equivalent system of linear DEs:
{y1′=a11y1+a12y2y2′=a21y1+a22y2\left\{\begin{array}{rcl} y_1'=a_{11}y_1+a_{12}y_2\\[0.5em] y_2'=a_{21}y_1+a_{22}y_2 \end{array}\right.

Wize Tip
We always want the left-hand side (LHS) to be a single derivative: y1′y_1' and y2′y_2'.

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Example
Write the equation y′′+3y′−y=0y''+3y'-y=0 as a system of linear differential equations.
Make the substitutions y1=yy_1=y and y2=y′y_2 = y'.
Equation 1
We want y1′y_1' on the LHS. Take the derivative of the first substitution: y1=y    ⟹    y1′=y′y_1=y\ \ \implies \ \ y_1'= \colorOne{y'}
We don't want any derivatives on the right-hand side (RHS), so we want to replace y′\colorOne{y'}.
Since the second substitution tells us y2=y′\colorTwo{y_2=y'}, we can write: y1′=y2y_1' = \colorTwo{y_2}.
So the first equation of our linear system is: y1′=y2\boxed{\quad y_1' = y_2\quad}
Equation 2
Now we are looking for y2′y_2', so we can take the derivative of our second substitution: y2=y′    ⟹    y2′=y′′y_2=y' \ \ \implies \ \ y_2' = \colorFour{y''}.
We don't want any derivatives on the right-hand side (RHS), so we want to replace y′′\colorFour{y''}.
Since we are given the equation y′′+3y′−y=0y''+3y'-y=0, we can solve for y′′y'':
y′′=y−3y′y''=y-3y'
Using the substitutions we made earlier (y1=yy_1=y and y2=y′y_2 = y'), we get the second equation of the system:
    y2′=y1−3y2    \boxed{\;\;y_2'=y_1-3y_2\;\;}
Then the system is:
{y1′=y2y2′=y1−3y2\left\{\begin{array}{rcl} y_1'&=&y_2\\[0.5em] y_2'&=&y_1-3y_2 \end{array}\right.
We can now solve this by writing it in matrix-vector form, as in the previous section.

Example: Higher Order Linear DEs

Solve the following second order differential equation if y(0)=2y(0)=2 and y′(0)=−2y'(0)=-2.

y′′+4y′−12y=0y''+4y'-12y=0

Let y1=yy_1=y and let y2=y′y_2=y'.
Then:
y1′=y′    ⟹    y1′=y2y_1'=y' \ \ \implies\ \ y_1'=y_2
And solving for y′′y'' in the original DE, we have y′′=12y−4y′y''=12y-4y', so we can write:
y2′=y′′    ⟹    y2′=12y−4y′    ⟹    y2′=12y1−4y2y_2' = y'' \ \ \implies \ \ y_2' = 12y-4y' \ \ \implies \ \ y_2'=12y_1-4y_2
So we have the linear system:
{y1′=y2y2′=12y1−4y2  ⟺  {y1′=0y1+1y2y2′=12y1−4y2\left\{\begin{array}{rcl} y_1'&=&y_2\\ y_2'&=&12y_1-4y_2 \end{array}\right. \quad \iff \quad \left\{\begin{array}{rcl} y_1'&=&0y_1 + 1y_2\\ y_2'&=&12y_1-4y_2 \end{array}\right.
Write this in matrix-vector form as:
[y1′y2′]=[0112−4]⏟A[y1y2]\left[\begin{array}{r} y_1'\\ y_2' \end{array}\right] = \underbrace{ \left[\begin{array}{rr} 0&1\\ 12&-4 \end{array}\right]}_{\normalsize A} \left[\begin{array}{r} y_1\\ y_2 \end{array}\right]
It can be shown that the eigenvalues and eigenvectors of AA are:
λ1=−6λ2=2x⃗1=[−16]x⃗2=[12]\begin{array}{l|l} \colorTwo{\lambda_1={-6}} \qquad& \quad \colorTwo{\lambda_2={2}}\\ \colorThree{\vec x_1=\begin{bmatrix}-1\\6 \end{bmatrix}} &\quad \colorThree{\vec x_2=\begin{bmatrix}1\\2 \end{bmatrix}} \end{array}
The eigenvalues are the coefficients of xx (in the exponent), and we multiply each term by the corresponding eigenvector:
y⃗=(C1eλ1x)x⃗1+(C2eλ2x)x⃗2  ⟹  [y1y2]=C1e−6x[−16]+C2e2x[12]\begin{aligned} \vec y &= (C_1e^{\colorTwo{\lambda_1} x})\colorThree{\vec x_1} +(C_2e^{\colorTwo{\lambda_2} x})\colorThree{\vec x_2}\\[1.5em] \implies \begin{bmatrix} y_1\\y_2\end{bmatrix} &= C_1e^{\colorTwo{-6} x}\colorThree{\begin{bmatrix} -1\\6 \end{bmatrix}} +C_2e^{\colorTwo{2} x}\colorThree{\begin{bmatrix} 1\\2 \end{bmatrix}} \end{aligned}
Multiplying into each vector yields the following solution (look at each row after multiplying):

{y1=−C1e−6x+C2e2xy2=6C1e−6x+2C2e2x\boxed{ \left\{\begin{array}{rllll} y_1&=&-C_1e^{-6x}&+C_2e^{2x}\\[0.5em] y_2&=&6C_1e^{-6x}&+2C_2e^{2x}\\[0.5em] \end{array}\right. }
We can now use the initial conditions to solve for C1C_1 and C2C_2.
Use the earlier substitions y1=yy_1=y and y2=y′y_2 = y', and recall the initial conditions:
  • y(0)=2    ⟹    y1(0)=2y(0)=2 \ \ \implies \ \ y_1(0)=2
  • y′(0)=−2    ⟹    y2(0)=−2y'(0)=-2 \ \ \implies \ \ y_2(0)=-2
This means we want y1=2y_1=2 and y2=−2y_2=-2 when x=0x=0.
Plugging these into the boxed solution:
{2=−C1e−6(0)+C2e2(0)=−C1+C2−2=6C1e−6(0)+2C2e2(0)=6C1+2C2\left\{\begin{array}{rllll} 2&=&-C_1e^{-6(0)}&+C_2e^{2(0)} &=& -C_1+C_2\\[0.5em] -2&=&6C_1e^{-6(0)}&+2C_2e^{2(0)} &=& 6C_1 + 2C_2\\[0.5em] \end{array}\right.
Solving (substitution, elimination, or row-reduction) yields: C1=−34,  C2=54C_1= -\dfrac{3}{4}, \ \ C_2=\dfrac{5}{4}
Note: the solution to the original DE is a single function yy. Our first substitution tells us this is simply y1y_1.
y1=−C1e−6x+C2e2xy_1 = -C_1e^{-6x}+C_2e^{2x}
  ⟹  y=−(−34)e−6x+54e2x\implies y = - \left(-\dfrac{3}{4}\right)e^{-6x}+\dfrac{5}{4}e^{2x}
  ⟹    y=34e−6x+54e2x  \implies \boxed{\;y=\frac{3}{4}e^{-6x}+\frac{5}{4}e^{2x}\;}

Practice: Higher Order DEs

Write the following differential equation as a linear system of differential equations. You do not need to solve the system.
y′′′+3y′′−8y′+2y=0y'''+3y''-8y'+2y=0
Which matrix AA appears in the equivalent system of linear differential equations?
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Practice: Higher Order DEs

Solve the following second order differential equation:
y′′−5y′+4y=0y''-5y'+4y=0
Extra Practice
For what value(s) of kk is y=ekxy=e^{kx} a solution to the second order differential equation 2y′′−5y′−3y=02y''-5y'-3y=0?
For what value(s) of kk is y=ekxy=e^{kx} a solution to the second order differential equation 2y′′−5y′−3y=02y''-5y'-3y=0?
For what value(s) of kk is y=ekxy=e^{kx} a solution to the second order differential equation 2y′′−5y′−3y=02y''-5y'-3y=0?
For what value(s) of kk is y=ekxy=e^{kx} a solution to the second order differential equation 2y′′−5y′−3y=02y''-5y'-3y=0?
Types of Differential Equations
Write down the order each of differential equation
a) y′′+3y′−8=0y''+3y'-8=0
b) d3Ndt3+100d2Ndt2=et\displaystyle\frac{d^3N}{dt^3}+100\frac{d^2N}{dt^2}=e^t