Question

3. Solve the differential equations a) y'' + 12y' + 32y = 0 b) y'' + 14y' + 49y = 0 c) y'' + 10y' + 34y = 0, y(0) = 1, y'(0) = 4

Answer 1

`y(x)=3e^(-4x)-2e^(-8x)`

Explanation:

I will do the first one thoroughly so you won't have any problems following to complete the rest of them.

This is a linear homogeneous second order differential, so to solve it we will use:

`y(x)=C_(1)e^{r_(1)x}+C_(2)e^{r_(2)x}` which is a theorem that says that if r1x and r2x are both solutions off a linear homogeneous equation, and C1 and C2 are any constants, then the function above is also a solution of the equation.

We need to solve for r1 and r2 using the differential equation:

y'' + 12y' + 32y = 0

Solve the differential equation for r1 and r2 by first replacing the y'' with r^2 and the y' with r:

`r^2+12r+32=0`

W will factor that now to solve for the 2 values of r:

(r + 4)(r + 8) = 0

By the Zero Product Property, either one of those binomials has to equal 0 for the product to equal 0, so

r + 4 = 0 and r = -4

r + 8 = 0 and r = -8

Those are the values for r1 and r2 and we can sub them back in to the y(x) equation:

`y(x)=C_(1)e^(-4x)+C_(2)e^(-8x)`

This we will call Equation 1.

Now we find the derivative of that equation, using the rules for finding derivatives of e's:

`y'(x)=-4C_(1)e^(-4x)-8C_(2)e^(-8x)`

This we will call Equation 2.

Now we will use our first initial condition in Equation 1, where y(0) = 1:

`y(0)=C_(1)e^((-4)(0))+C_(2)e^((-8)(0))=1`

Simplifying gives you:

`y(0)=C_(1)e^0+C_(2)e^0=1` so

`C_(1)+C_(2)=1`

Now we will use the second initial condition in Equation 2, where y'(0) = 4:

`y'(0)=-4C_(1)e^((-4)(0))-8C_(2)e^((-8)(0))=4`

Simplifying gives you:

`y'(0)=-4C_(1)e^0-8C_(2)e^0=4` so

`-4C_(1)-8C_(2)=4`

We will now go back to the first bold equation and solve it for C1:

`C_(1)=1-C_(2)` and sub that value in to the second bold equation to solve for C2:

`-4(1-C_(2))-8C_(2)=4` and

`-4+4C_(2)-8C_(2)=4` and

`-4C_(2)=8` so

`C_(2)=-2`

Now sub that back in to the first bold equation to solve for C1:

`C_(1)-2=1` so

`C_(1)=3`

Finally we go back to the y(x) equation and fill everything in:

`y(x)=3e^(-4x)-2e^(-8x)`

And that's your original equation! Follow this to the "t" and you'll have no problems with the other 2. They are identical in execution.