Solving a Second-Order Differential Equation: y" - 3y' - 10y = 1

Published on May 29, 2025

Calculus

Find the solution y(t) to the second-order linear non-homogeneous differential equation y" - 3y' - 10y = 1 with initial conditions y(0) = -1 and y'(0) = 2. The solution involves finding the complementary and particular solutions and applying initial conditions.

Question

Answer

The problem asks us to find the solution

to the given initial value problem (IVP):

with initial conditions

and

This is a second-order linear non-homogeneous differential equation with constant coefficients. The general solution

can be found by summing the complementary solution

and a particular solution

, i.e.,

Step 1: Find the complementary solution

.
The homogeneous equation is

.
The characteristic equation is

.
We can factor this quadratic equation:

The roots are

and

.
Since the roots are real and distinct, the complementary solution is:

where

and

are arbitrary constants.

Step 2: Find a particular solution

.
Since the right-hand side of the non-homogeneous equation is a constant,

, we can assume a particular solution of the form

, where

is a constant.
Then,

and

.
Substitute these into the non-homogeneous equation:

So, the particular solution is

Step 3: Form the general solution.
The general solution is

Step 4: Apply the initial conditions to find

and

.
We have

and

.
First, let’s find

Now, apply the initial conditions:
For

For

Now we have a system of two linear equations with two variables

and

From equation (1),

. Substitute this into equation (2):

Now substitute

back into

To subtract these fractions, find a common denominator, which is 70:

Step 5: Write the final solution

.
Substitute the values of

and

into the general solution:

The final answer is

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Solving a Second-Order Differential Equation: y" - 3y' - 10y = 1

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