plzz solve diff equationsSolutionGiven differential equation

plzz solve diff equations

Solution

Given differential equation can be written as:

y\'\' = - p y\' - qy,

where p(t) and q(t) are known functions of t

and y(t) is the yet to be determined function.

Given; y₁ and y₂ are solutions of the differential equation.

Then, the Wronskian is given by:

W(t) = y₁\' y₂ - y₂\' y_1..

(b) The given differential equation Second Order Homogeneous Linear Differential Equation.

For proving y=C₁ y₁ + C₂ y₂ is also a solution to the differential equation, first prove the following theorem:

Given: y\'\' + p y + q y = 0                            (1)

Let y = y₁ (t) be a solution of (1).

Then: u(t) = C₁ y₁ (t), where C₁ is any real number is also a solution of (1).

Proof:

Let y₁ = y₁ (t) be a solution of (1).

Then,

y\'\' (t) + p(t) y\'(t) + q(t) y(t) =0.

Let C₁ be any real number and set:

u (t) = C₁ y(t),

u\'(t) = C₁ y\'(t).

u\'\'(t) = C₁ y\'\'(t).

Substituting u in (1), we get:

u\'\'(t) + p(t) u\'(t) + q(t) u(t)

= C₁ y\'\'(t) + p(t) (C₁ y\'(t)) + q(t) (C₁ y(t))

= C₁ (y\'\'(t) + p(t) y\'(t) +q(t) y(t))

= C₁ (0)

= 0.

Similary, we can prove :

v(t) + C₂ y₂(t) is also a solution.

Thus, summing up these two solution,

y = C₁ y₁ + C₂ y₂ is a solution of the differential equation.This is by the Principle of Superposion.

(c) If the Wronskian is never zero, It implies that y₁ and y₂ are linearly independent.

By Existance and Uniqueness Theorem on Solution of Linear Homogeneous Differential Equation, if the Wronskian is not 0,

y=C₁ y₁ + C₂ y₂ forms a Fundamental Set of Solutions and as a consequence, for any initial conditions y(0)=A and y\'(0) = B, there is a unique set of (C₁, C₂) that give a unique solution. This proves the result (c).