Solve the given differential equation by undetermined coefficients. y(4) + 2y'' + y = (x - 5)²

Solution:

Given, the differential equation is y(4) + 2y'' + y = (x - 5)²

The differential operator form of the given differential equation is

(D⁴ - 2D³ + D²)y = e^x + 1

Comparing f(d)y = e^x + 1

The auxiliary equation f(m) = 0

m⁴ - 2m³ + m² = 0

m²(m²- 2m + 1) = 0

Now, m² - 2m + 1 = m² - m - m + 1

= m(m - 1) - 1(m - 1)

= (m - 1)(m - 1)

So, m²(m - 1)(m - 1) = 0

m² = 0

m = ±0

m - 1 = 0

m = 1

The roots are 0, 0, 1 and 1.

Complementary function is \(y_{c}=(C_{1}+C_{1}x)e^{0}x+(C_{3}+C_{4}x)e^{x}\)

The particular equation is \(\frac{1}{f(D)}Q\)

P.I = \(\frac{1}{D^{2}(D-1)^{2}}e^{x}+1\)

P.I = \(\frac{1}{D^{2}(D-1)^{2}}e^{x}+\frac{1}{D^{2}(D-1)^{2}}e^{0}\)

P.I = I₁ + I₂

= \(\frac{1}{D^{2}}(\frac{x^{2}}{2!})e^{x}+\frac{1}{D^{2}}e^{0x}\)

1/D means integration,

\(\frac{1}{D^{2}}(\frac{x^{2}}{2!})e^{x}=\frac{1}{2D}\int x^{2}e^{x}dx\)

Applying uv formula,

\(\int uv\, dx=u\int v\, dx\, -\, \int (u^{l}\int v\, dx)dx\)

\(\\I_{1}=\frac{1}{D^{2}(D-1)^{2}}e^{x}\\=\frac{1}{2D}(e^{x}(x^{2}-e^{x}(2x)+e^{x}(2)))\\=\frac{1}{2}(e^{x}(x^{2}-2x+2)-e^{x}(2(x-1)+e^{x}(2)))\)

\(\\I_{2}=\frac{1}{D^{2}(D-1)^{2}}e^{0x}\\=\frac{1}{D}\int 1\, dx\\=\frac{1}{D}x\)

Again integrating, \(\frac{1}{D}x = \frac{x^{2}}{2!}\)

The general solution is y = \(y_{c}+y_{p}\)

Therefore,\(y=(C_{1}+C_{1}x)e^{0}x+(C_{3}+C_{4}x)e^{x}+\frac{1}{2}(e^{x}(x^{2}-2x+2)-e^{x}(2(x-1)+e^{x}(2)))+\frac{x^{2}}{2!}\).