Find the general solution y'' - 3y' + 2y = 21

Question

How would I find the general solution of y'' − 3y' + 2y = 21 ?

Answer 1

This is a non homogeneous equation:

y'' + p(t)y' +q(t)y = g(t), where g(t) ≠ 0

 

So the general solution to the non homogeneous equation:

y(t)= c₁y₁ + c₂y₂ + Y(t), where Y is any one particular solution to the homogeneous equation.

 

Step 1: Find the general solution to the homogeneous equation so you set the equation equal to zero:

 

y'' - 3y' + 2y = 0, take the coefficients of y'', y' and y and place them in ar² +br+c=0

 

r² - 3r +2 = 0, now factor

 

(r-2)(r-1), so r = 1, 2

 

The general solution for the homogeneous equation is:

 

c₁e^t+c₂e^2t

 

Step 2: Find one particular solution Y(t) to the non homogeneous equation:

 

y'' - 3y' + 2y = 21, set y=A because 21 is just a constant without an x, if the equation equaled t² then y = (At²+Bt+C)

 

if y=A, y'=0 and y''=0

 

Plug in y, y' and y'' and you get

 

2y=21, solve for y and you get 

 

Y(t)=21/2

 

Step 3: put it all together

 

y(t)= c₁y₁ + c₂y₂ + Y(t)

 

The general solution:

 

y(t)= c₁e^t+c₂e^2t+ (21/2)

Answer 2

We have here a non-homogenous second order differential equation because their is a second derivative term and the equation is NOT equal to zero.

 

First is find the general solution when the equation is homogenous.  Then we find the general solution when the equation is non-homogenous. Then we combine the two general solutions.

 

Think of this equation as y'' - y' + 2y = 0 + 21.

 

Step 1: Solve for homogenous part.

 

Let y = e^rt

 

then y' = re^rt

then y'' = r²e^rt

 

 

Substitute these values into the homogenous equation.

 

r²e^rt - 3re^rt + 2e^rt = 0

 

Factor out the common terms on the left side of equation.

 

e^rt(r² - 3r + 2) = 0

 

Notice that when we set the term in parenthesis equal to zero, it makes both sides equal to zero.  We use this term as our characteristic equation.

 

r² - 3r + 2 = 0

(r - 2)(r  - 1) = 0

r = 2  ,   r = 1

 

The general solution of a homogenous is in the form

 

y = C1e^r₁t + C₂e^r₂t   ,    when the roots are different

y = C₁e^rt + C₂te^rt    ,    when the roots repeat themselves

y = C₁e^atcos(bt) + C₂e^atsin(bt) , when we have complex roots  a ± bi

 

where C₁ and C₂ are constants. If we had initial value conditions, then we would have solved for C₁ and C₂. 

 

We have two different roots.  Our general solution here is

 

y_H = C₁e^2t + C₂e^t

 

 

Step 2: Solve for non-homogenous part.

 

The right side of the equation is a constant.  In that case, we must let y equal a function that acts just like a constant.

 

Let y = A

 

A is some constant.

 

then y' = 0

then y'' = 0

 

Substitute these values into the non-homogenous equation.

 

0 - 0 + 2A = 21

           2A = 21

             A = 21/2

 

Since y = A , then the general solution to this non-homogenous part of the equation is

 

y_P = 21/2

 

 

Next, combine the two solutions.  The final solution is

 

y = y_H + y_P

y = C₁e^2t + C₂e^t + (21 / 2)