y^{''}-2y^'+5y=5x+3 — step-by-step solution

1) Let's solve the differential equation

y'' - 2 y' + 5 y = 5 x + 3

, where

y

is a function of

x

For the homogeneous equation, we look for exp(r*

x

): the characteristic equation is

r^{2} - 2 r + 5 = 0

2) Let's solve in

ℂ

r^{2} - 2 r + 5 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot 5

Δ = 4 - 4 \cdot 5

Δ = 4 - 20

Δ = -16

Δ < 0

, the equation has two solutions:

Solution first:

\frac{- b - \sqrt{Δ}}{2 a}

r1 = \frac{2 - \sqrt{-16}}{2}

r1 = \frac{2 - i \cdot 4}{2}

r1 = 1 - 2 i

For the second solution, we obtain:

\frac{- b + \sqrt{Δ}}{2 a}

r2 = \frac{2 + \sqrt{-16}}{2}

r2 = \frac{2 + i \cdot 4}{2}

r2 = 1 + 2 i

Solution found:

r = 1 - 2 i

Solution found:

r = 1 + 2 i

The homogeneous solution is

\exp (x) \cdot C1 \cdot \cos (2 x) + \exp (x) \cdot C2 \cdot \sin (2 x)

We look for a particular solution A*

x

+ B. Its derivatives are A and 0, so

5 A = 5

and

- 2 A + 5 B = 3

We obtain A =

1

and B =

1

, so the particular solution is

x + 1

The general solution is the sum of the homogeneous and particular solutions;

C1

and

C2

are arbitrary real constants

▶Solution:

y = \exp (x) \cdot C1 \cdot \cos (2 x)

+ \exp (x) \cdot C2 \cdot \sin (2 x)

+ x

+ 1