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:[tex]y(x)=3e^{-4x}-2e^{-8x}[/tex]Step-by-step 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:[tex]y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}[/tex] 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 = 0Solve the differential equation for r1 and r2 by first replacing the y'' with r^2 and the y' with r:[tex]r^2+12r+32=0[/tex]W will factor that now to solve for the 2 values of r:(r + 4)(r + 8) = 0By the Zero Product Property, either one of those binomials has to equal 0 for the product to equal 0, sor + 4 = 0 and r = -4r + 8 = 0 and r = -8Those are the values for r1 and r2 and we can sub them back in to the y(x) equation:[tex]y(x)=C_{1}e^{-4x}+C_{2}e^{-8x}[/tex]This we will call Equation 1.Now we find the derivative of that equation, using the rules for finding derivatives of e's:[tex]y'(x)=-4C_{1}e^{-4x}-8C_{2}e^{-8x}[/tex]This we will call Equation 2.Now we will use our first initial condition in Equation 1, where y(0) = 1:[tex]y(0)=C_{1}e^{(-4)(0)}+C_{2}e^{(-8)(0)}=1[/tex]Simplifying gives you:[tex]y(0)=C_{1}e^0+C_{2}e^0=1[/tex] so[tex]C_{1}+C_{2}=1[/tex]Now we will use the second initial condition in Equation 2, where y'(0) = 4:[tex]y'(0)=-4C_{1}e^{(-4)(0)}-8C_{2}e^{(-8)(0)}=4[/tex]Simplifying gives you:[tex]y'(0)=-4C_{1}e^0-8C_{2}e^0=4[/tex] so[tex]-4C_{1}-8C_{2}=4[/tex]We will now go back to the first bold equation and solve it for C1:[tex]C_{1}=1-C_{2}[/tex] and sub that value in to the second bold equation to solve for C2:[tex]-4(1-C_{2})-8C_{2}=4[/tex] and[tex]-4+4C_{2}-8C_{2}=4[/tex] and[tex]-4C_{2}=8[/tex] so[tex]C_{2}=-2[/tex]Now sub that back in to the first bold equation to solve for C1:[tex]C_{1}-2=1[/tex] so[tex]C_{1}=3[/tex]Finally we go back to the y(x) equation and fill everything in:[tex]y(x)=3e^{-4x}-2e^{-8x}[/tex]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.