General solution of the following DEs

Sample Solution

I'd be glad to help you obtain the general solutions of the given differential equations:

i. y''' + y'' - 4y' + 2y = 0

1. Characteristic Equation: Assume a solution of the form y = e^(mx). Substitute this into the equation: m^3 + m^2 - 4m + 2 = 0 Factor the equation: (m - 1)(m^2 + 2m - 2) = 0

2. Roots of the Characteristic Equation:

   • m1 = 1 (repeated root)
   • For m^2 + 2m - 2 = 0, use the quadratic formula: m2,3 = (-2 ± sqrt(2^2 - 4 * 1 * -2)) / (2 * 1) m2,3 = (-2 ± 2sqrt(2)) / 2 m2 = -1 + sqrt(2) m3 = -1 - sqrt(2)

3. General Solution: Since there's a repeated root (m1), the general solution will have the form: y(x) = C1e^(m1x) + C2xe^(m1x) + C3e^(m2x) + C4e^(m3*x)

   where C1, C2, C3, and C4 are arbitrary constants determined by initial conditions.

ii. y(4) + 4y(2) = 0

This equation is not a homogeneous differential equation with constant coefficients. It appears to be an initial value problem (IVP) with specific initial conditions at x = 2 and x = 4. To solve IVPs, we typically need the complete initial conditions (values of y and its derivatives at specific points). Without them, we cannot determine a unique solution.

iii. x(x - 2)y'' + 2(x - 1)y' - 2y = 0; use y1 = (1 - x)

This equation appears to have a non-constant coefficient (x(x - 2) in the second derivative term). It's generally more challenging to solve such equations directly. However, if y1 = (1 - x) is a solution, we can use reduction of order to find a second solution (y2) and construct the general solution:

1. Assume a second solution of the form y2 = v(x)y1(x) = v(x)(1 - x).
2. Substitute y2 into the equation and solve for v(x).
3. The general solution will be y(x) = C1y1(x) + C2y2(x) = C1*(1 - x) + C2v(x)(1 - x).

iv. y'' - 4y = sin^2(x)

Full Answer Section

1. Characteristic Equation: m^2 - 4 = 0 (m - 2)(m + 2) = 0

2. Roots of the Characteristic Equation:

   • m1 = 2
   • m2 = -2

3. General Solution: y(x) = C1e^(2x) + C2e^(-2x) + Particular Solution (for sin^2(x) term)

   The particular solution for sin^2(x) can be found using methods like undetermined coefficients or variation of parameters.

v. y'' - 4y' + 3y = x ; use y1 = e^3x

1. Characteristic Equation: m^2 - 4m + 3 = 0 (m - 1)(m - 3) = 0

2. Roots of the Characteristic Equation:

   • m1 = 1
   • m2 = 3

3. General Solution (using y1 = e^3x): y(x) = C1e^(x) + C2e^(3x) + Particular Solution (for x term)

   The particular solution for x can be found using methods like undetermined coefficients.

vi. y'' + 5y' + 6y = e^2xcos(x)

1. Characteristic Equation: m^2 + 5m + 6 = 0 (m + 2)(m + 3) = 0

2. Roots of the Characteristic Equation:

   • m1 = -2
   • m2 = -3

3. General Solution: y