Step by step solution to "Determine the general solution of the given differential..."

Oliver Tyler

Answered question

2022-03-28

Determine the general solution of the given differential equation:

y^{″} - 4 y^{'} + 13 y = e^{2 x}

allemelrypi0e

Beginner2022-03-29Added 11 answers

Step 1

y (x) = y_{h} (x) + y_{p} (x),

where

y_{h} (x)

is a solution of a homogeneous equation

y^{″} - 4 y^{'} + 13 y = 0

, and

y_{p} (x)

is a particular solution.
To solve the equation

y^{″} - 4 y^{'} + 13 y = 0

, one have to solve a characteristic equation

λ^{2} - 4 λ + 13 = 0,

which is a quadratic algebraic equation. The solution is

λ_{1, 2} = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 \cdot 1 \cdot 13}}{2 \cdot 1} = 2 \pm 3 i \Rightarrow

y_{h} (x) = C_{1} e^{2 x} \cos (3 x) + C_{2} e^{2 x} \sin (3 x)

.
Step 2
The solution

y_{p} (x)

can be found in a form

y_{p} (x) = A e^{2 x}

. Then,

y_{p}^{'} (x) = 2 A e^{2 x}, and y_{p}^{″} (x) = 4 A e^{2 x} .

Substituting

y_{p} (x), y_{p}^{'} (x), and y_{p}^{″} (x)

into the equation, one has

4 A e^{2 x} - 4 \cdot 2 A e^{2 x} + 13 A e^{2 x} = e^{2 x} \Leftrightarrow 9 A = 1 \Leftrightarrow A = \frac{1}{9},

and

y_{p} (x) = \frac{1}{9} e^{2 x}

Finally, the general solution is

y (x) = y_{h} (x) + y_{p} (x) = C_{1} e^{2 x} \cos (3 x) + C_{2} e^{2 x} \sin (3 x) + \frac{1}{9} e^{2 x}

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