Problems Based On Homogeneous Equations

Problems based on Homogeneous equations

Example 1.5.1: Solve (D² - 4DD' + 3D'²) z = 0.

Solution: Given [D² - 4DD' + 3D'²] z = 0

The auxiliary equation is m² - 4m + 3 = 0

[Replace D by m and D' by 1]

Solving, m² - 3m -m +3 = 0

Since, R.H.S. is zero, there is no particular integral

Hence, the general solution is z = C.F.

+ϕ 1 (y + x) + 2 (v + 3x)

Example 1.5.2: Solve [D² - 2DD' + D'²] z = 0.

Solution: [D² - 2DD' + D'²]z = 0.

The auxiliary equation is m² - 2m + 1 = 0

[Replace D by m and D' by 1]

i.e., (m - 1)² = 0

m = 1, 1

Here, the roots are equal

.. C.F. = ϕ₁( v+x) + xϕ ₂ (v + x).

Since R.H.S. is zero, there is no particular integral

Hence, the general solution is z = C.F

Z = ϕ₁(y+x) + xϕ ₂ (y + x).

Example 1.5.3: Solve [D³ + DD² - D2 D' – D'³] z = 0

Solution : Given [D³ + DD² - D2 D' – D'³] z = 0

The auxiliary equation is m3 - m2 + m - - 1 = 0

 [Replace D by m and D' by 1] a

m² (m-1) + (m − 1) = 0

 (m − 1) (m² + 1) = 0

m = 1, m² + 1 = 0

i.e., m = 1, m =  ±i

i.e., m = 1, m = i, m = -i

Here, the roots are distinct

.. C.F = ϕ₁ (y + x) + = 0₂ (y+ix) + = 0₃ (v-ix)

Since R.H.S. is zero, there is no particular integral

Hence, the general solution is

Z = ϕ₁ (y + x) + = 0₂ (y+ix) + = 0₃ (v-ix)

Example 1.5.4: Solve 2r+ 5s - 3t = 0.

Solution: The given differential equation can be written as

The auxiliary equation is 2m² + 5m - 3 = 0

[Replace D by m and D' by 1]

2m² + 6m-m-3 = 0

2m (m + 3) - 1 (m + 3) = 0

 (m + 3) (2m −1) = 0

m= -3, m=1/2

Here the roots are distinct.

Since R.H.S. is zero, there is no particular integral

Hence, the general solution is

Example 1.5.5 : Solve (D⁴ - D4) z = 0

Solution : Given (D⁴— D'⁴) z = 0

The auxiliary equation is ppppp

[Replace D by m and D' by 1]