Question

solve the second order differential equation d^2y/dx^2-4y=12x, y(0)=4, y'(0)=1

Answer 1

Answer to Question #89431 – Math – Differential Equations

Question

Solve the second order differential equation

$$d^2y/dx^2 - 4y = 12x$$

,

$$y(0) = 4$$

,

y'(0) = 1$ ## Solution Let

y = p + q

$$where$$

p

$$and$$

q

$$are the function of$$

x

$$, then$$

y'' = p'' + q''

$$So,$$

p'' + q'' - 4p - 4q = 12x

$$If$$

p'' - 4p = 0

$$can be solved we only have to solve$$

q'' - 4q = 12xp'' - 4p = 0

$$has the characteristics equation$$

(m - 2)(m + 2) = 0

$$and we use the roots to suggest what$$

p

$$might be:$$

\begin{aligned}

p &= A e^{\wedge}(2x) + B e^{\wedge}(-2x); \\

p' &= 2 A e^{\wedge}(2x) - 2 B e^{\wedge}(-2x) \\

p'' &= 4 A e^{\wedge}(2x) + 4 B e^{\wedge}(-2x) = 4p.

\end{aligned}

$$Now, for$$

q

$$. Since the right-hand side of the differential equation is$$

12x

$$,$$

q = ax + b

$$and$$

q'' = 0

$$,$$

q' = a

$$, a=constant;$$

\begin{aligned}

q &= ax + b \\

q' &= a \\

q'' &= 0 \quad \text{(Derivative of the constant is equal to zero)}

\end{aligned}

q'' - 4q

$$must be equal to$$

12x

$$so$$

-4ax - 4b = 12x

$$, making$$

b = 0

$$and$$

a = -3

$$Therefore,$$

\mathrm {q} = - 3 \mathrm {x}

y = p + q = A e ^ {\wedge} (2 x) + B e ^ {\wedge} (- 2 x) - 3 x

y ^ {\prime} = 2 A e ^ {\wedge} (2 x) - 2 B e ^ {\wedge} (- 2 x) - 3

$$After putting values,$$

y (0) = 4 = A + B

y ^ {\prime} (0) = 1 = 2 A - 2 B - 3

$$So, we have$$

\mathrm {A} = 3 \text { and } \mathrm {B} = 1

Y = 3 e ^ {\wedge} (2 x) + e ^ {\wedge} (- 2 x) - 3 x

$$

Answer: $Y = 3e^{\wedge}(2x) + e^{\wedge}(-2x) - 3x$ .

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