# Solving Forced Equations: y'' + 2y' = 3t+2

• KevinL
In summary: The specific solution is y(t) = k_1e^{-2t}+ K_2e^{0t}. The general solution is y(t) = k_1e^{-2t}+ K_2.
KevinL
I need to compute the general solution to:

y'' + 2y' = 3t+2

The only method we have been taught is to first find the general solution of the unforced equation and then find the particular solution.

For general solution, the characteristic polynomial is L^2 + 2L = 0 (where L= lambda)

L^2 = -2L
L = -2

So general solution of unforced equation is y(t) = k1e^-2t +K2te^-2t

I am fairly certain about the above. The particular solution is a bit trickier for me.

guess: yp(t) = At + b

Plug in: 2A = 3t + 2
A = (3/2)t + 1

Is that correct? It seems like an odd answer, and B sort of just disappeared.

Don't forget that the general solution to the NH equation consists of the general solution to the homogeneous equation plus a particular solution to the NH equation.

KevinL said:
I need to compute the general solution to:

y'' + 2y' = 3t+2

The only method we have been taught is to first find the general solution of the unforced equation and then find the particular solution.

For general solution, the characteristic polynomial is L^2 + 2L = 0 (where L= lambda)

L^2 = -2L
L = -2
Well, no. You get L= -2 if you divide by L which you cannot do if L= 0. The two roots are L= -2 and L= 0.

So general solution of unforced equation is y(t) = k1e^-2t +K2te^-2t
No, the general solution to the unforced equation is $y(t)= k_1e^{-2t}+ K_2e^{0t}$$= k_1e^{-2t}+ K_2$.

I am fairly certain about the above. The particular solution is a bit trickier for me.

guess: yp(t) = At + b

Plug in: 2A = 3t + 2
A = (3/2)t + 1

Is that correct? It seems like an odd answer, and B sort of just disappeared.
No that is not correct. The whole point of writing "y= At+ B" is that A and B are to be constants. A cannot be "(3/2)t+ 1" so your particular solution is NOT "(3/2)t^2+ 2t".

If you set y= At+ B, then y'= A and y"= B. That means your equation becomes y"+ 2y'= 0+ B= 3t+ 2. That's impossible because we have no "t" on the left. The problem is that the "3t+ 2" correspond to multiples of $e^{0t}= 1$ which is already a solution to the associated homogeneous equation. Just like with a multiple root, you should try multiplying by "t". If $y(t)= At^2+ Bt$, then $y'= 2At+ B$ and $y"= 2A$. Now the equation becomes $y"+ 2y= 2A+ 4At+ 2B= 4At+ (2A+ 2B)= 2t+ 3$. For that to be true for all t, we must have 4A= 2 and 2A+ 2B= 3. From the first equation, A= 1/2. Then the second equation becomes 1+ 2B= 3 so B= 1.

Now, write down the specific solution and the general solution to the entire equation.

## What is a forced equation?

A forced equation is a type of differential equation that includes an external force or input term. In the context of solving equations, it refers to finding a particular solution to the equation that satisfies the given external force or input.

## How is a forced equation different from a regular differential equation?

A regular differential equation does not have an external force term, while a forced equation does. This external force term can be a constant or a function of the independent variable.

## What is the general approach to solving forced equations?

The general approach to solving forced equations is to first find the complementary solution, which is the solution to the equation without the external force term. Then, a particular solution is found by using a method called variation of parameters, which involves finding a function that satisfies the external force term and plugging it into the original equation.

## Can a forced equation have multiple solutions?

Yes, a forced equation can have multiple solutions. The complementary solution is always unique, but the particular solution can vary depending on the chosen function for variation of parameters.

## What are some applications of solving forced equations in real life?

Forced equations are commonly used in physics and engineering to model systems that are affected by external forces. Examples include the motion of a mass on a spring under the influence of an external force, or the behavior of an electrical circuit with a varying input voltage.

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