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7.1 Second-order linear equations  (Page 6/15)

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y ( x ) = c 1 e ( α + β i ) x + c 2 e ( α − β i ) x .

Using some smart choices for c 1 and c 2 , and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to [link] and express our general solution in those terms.

We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was Euler’s formula , which tells us that

e i θ = cos θ + i sin θ

for all real numbers θ .

Going back to the general solution, we have

y ( x ) = c 1 e ( α + β i ) x + c 2 e ( α − β i ) x = c 1 e α x e β i x + c 2 e α x e − β i x = e α x ( c 1 e β i x + c 2 e − β i x ) .

Applying Euler’s formula together with the identities cos ( − x ) = cos x and sin ( − x ) = − sin x , we get

y ( x ) = e α x [ c 1 ( cos β x + i sin β x ) + c 2 ( cos ( − β x ) + i sin ( − β x ) ) ] = e α x [ ( c 1 + c 2 ) cos β x + ( c 1 − c 2 ) i sin β x ] .

Now, if we choose c 1 = c 2 = 1 2 , the second term is zero and we get

y ( x ) = e α x cos β x

as a real-value solution to [link] . Similarly, if we choose c 1 = − i 2 and c 2 = i 2 , the first term is zero and we get

y ( x ) = e α x sin β x

as a second, linearly independent, real-value solution to [link] .

Based on this, we see that if the characteristic equation has complex conjugate roots α ± β i , then the general solution to [link] is given by

y ( x ) = c 1 e α x cos β x + c 2 e α x sin β x = e α x ( c 1 cos β x + c 2 sin β x ) ,

where c 1 and c 2 are constants.

For example, the differential equation y ″ − 2 y ′ + 5 y = 0 has the associated characteristic equation λ 2 − 2 λ + 5 = 0 . By the quadratic formula, the roots of the characteristic equation are 1 ± 2 i . Therefore, the general solution to this differential equation is

y ( x ) = e x ( c 1 cos 2 x + c 2 sin 2 x ) .

Summary of results

We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in [link] .

Summary of characteristic equation cases
Characteristic Equation Roots General Solution to the Differential Equation
Distinct real roots, λ 1 and λ 2 y ( x ) = c 1 e λ 1 x + c 2 e λ 2 x
A repeated real root, λ y ( x ) = c 1 e λ x + c 2 x e λ x
Complex conjugate roots α ± β i y ( x ) = e α x ( c 1 cos β x + c 2 sin β x )

Problem-solving strategy: using the characteristic equation to solve second-order differential equations with constant coefficients

  1. Write the differential equation in the form a y ″ + b y ′ + c y = 0 .
  2. Find the corresponding characteristic equation a λ 2 + b λ + c = 0 .
  3. Either factor the characteristic equation or use the quadratic formula to find the roots.
  4. Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.

Solving second-order equations with constant coefficients

Find the general solution to the following differential equations. Give your answers as functions of x .

  1. y ″ + 3 y ′ − 4 y = 0
  2. y ″ + 6 y ′ + 13 y = 0
  3. y ″ + 2 y ′ + y = 0
  4. y ″ − 5 y ′ = 0
  5. y ″ − 16 y = 0
  6. y ″ + 16 y = 0

Note that all these equations are already given in standard form (step 1).

  1. The characteristic equation is λ 2 + 3 λ − 4 = 0 (step 2). This factors into ( λ + 4 ) ( λ − 1 ) = 0 , so the roots of the characteristic equation are λ 1 = −4 and λ 2 = 1 (step 3). Then the general solution to the differential equation is
    y ( x ) = c 1 e −4 x + c 2 e x (step 4) .
  2. The characteristic equation is λ 2 + 6 λ + 13 = 0 (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots −3 ± 2 i (step 3). Then the general solution to the differential equation is
    y ( t ) = e −3 t ( c 1 cos 2 t + c 2 sin 2 t ) (step 4) .
  3. The characteristic equation is λ 2 + 2 λ + 1 = 0 (step 2). This factors into ( λ + 1 ) 2 = 0 , so the characteristic equation has a repeated real root λ = −1 (step 3). Then the general solution to the differential equation is
    y ( t ) = c 1 e − t + c 2 t e − t (step 4).
  4. The characteristic equation is λ 2 − 5 λ (step 2). This factors into λ ( λ − 5 ) = 0 , so the roots of the characteristic equation are λ 1 = 0 and λ 2 = 5 (step 3). Note that e 0 x = e 0 = 1 , so our first solution is just a constant. Then the general solution to the differential equation is
    y ( x ) = c 1 + c 2 e 5 x (step 4) .
  5. The characteristic equation is λ 2 − 16 = 0 (step 2). This factors into ( λ + 4 ) ( λ − 4 ) = 0 , so the roots of the characteristic equation are λ 1 = 4 and λ 2 = −4 (step 3). Then the general solution to the differential equation is
    y ( x ) = c 1 e 4 x + c 2 e −4 x (step 4) .
  6. The characteristic equation is λ 2 + 16 = 0 (step 2). This has complex conjugate roots ± 4 i (step 3). Note that e 0 x = e 0 = 1 , so the exponential term in our solution is just a constant. Then the general solution to the differential equation is
    y ( t ) = c 1 cos 4 t + c 2 sin 4 t (step 4) .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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