2nd order differential equations

[pyroknife]

y''-3y'+2y=e^t y(0)=0 y'(0)=-1

yh=solution to homogeneous equation (y''-3y'+2y=0) = Ce^t+Ae^(2t)
C and A are constants

yp=solution to particular solution (e^t)
yp=ae^t where a is a constant. It turns out that this solves the homogenuous solution so I had to multiply it by a factor of t giving
yp=a*t*e^t
yp'=a(e^t+te^t)
yp''=a(e^t+e^t+te^2)=a(2e^2+te^2)
subsitution this into the original equation gives a=-1. Thus yp=-te^t

the solution is then y(t)=yh+yp=Ce^t+Ae^(2t)-te^t
Now I solve for C and A

0=C+A C=-A
y'(t)=Ce^t+2Ae^(2t)-e^t-te^2
-1=C+2A-1 -1=-A+2A-1 A=0 C=0


The problem with this is when I solve it I get that C=A=0 which means that the homogenuous solution was a TRIVIAL solution. I've never experienced a problem like this yet. Would that just mean the solution is y=yp=-te^2?

 

[Mark44]

y''-3y'+2y=e^t y(0)=0 y'(0)=-1

yh=solution to homogeneous equation (y''-3y'+2y=0) = Ce^t+Ae^(2t)
C and A are constants

yp=solution to particular solution (e^t)
yp=ae^t where a is a constant. It turns out that this solves the homogenuous solution so I had to multiply it by a factor of t giving
yp=a*t*e^t
yp'=a(e^t+te^t)
yp''=a(e^t+e^t+te^2)=a(2e^2+te^2)
subsitution this into the original equation gives a=-1. Thus yp=-te^t

the solution is then y(t)=yh+yp=Ce^t+Ae^(2t)-te^t
Now I solve for C and A

0=C+A C=-A
y'(t)=Ce^t+2Ae^(2t)-e^t-te^2
-1=C+2A-1 -1=-A+2A-1 A=0 C=0


The problem with this is when I solve it I get that C=A=0 which means that the homogenuous solution was a TRIVIAL solution. I've never experienced a problem like this yet. Would that just mean the solution is y=yp=-te^2?

You can check for yourself.
First, verify that your general solution satisifies the initial conditions.
Second, verify that your general solution satisfies the differential equation.

If both check out, you're golden.

 

[pyroknife]

You can check for yourself.
First, verify that your general solution satisifies the initial conditions.
Second, verify that your general solution satisfies the differential equation.

If both check out, you're golden.

Yes, I know that, but in this case the particular solution is also the general solution. The particular solution always solves the differential equation and its initial conditions. Checking that would only prove the particular solution solves the diff eq. If I made a mistake where the homogenuous equation was not = 0, then this method will not let me see that.

 

[Mark44]

That all comes out in the wash, so to speak. Again, if your general solution satisfies the initial conditions and the differential equation, it's good.

The complementary function, which is the solution to the homogeneous equation, is a solution to the homogeneous equation for any values of the constants, including the case where both are zero. For your problem, the complementary function was the trivial solution because of the initial conditions.