Homogeneous differential equations initial value problems
Process for solving an initial value problem for a second-order homogeneous differential equation
We’ve already learned how to find the general solution of a second-order homogeneous differential equation.
The general solutions will look something like these:
Even though we’ve plugged in for the roots of the differential equation, the constants ???c_1??? and ???c_2??? remain in the general solution. The only way to solve for these constants is with initial conditions.
In a second-order homogeneous differential equations initial value problem, we’ll usually be given one initial condition for the general solution, and a second initial condition for the derivative of the general solution. We’ll apply the first initial condition to the general solution to get a simple equation in terms of ???c_1??? and ???c_2???. Then we’ll take the derivative of the general solution so that we can apply the second initial condition to the derivative. This will give us a second simple equation in terms of ???c_1??? and ???c_2???.
Once we have two equations in terms of ???c_1??? and ???c_2???, we can solve them as a system of linear equations, find the values of ???c_1??? and ???c_2???, and then plug these values back into the general solution.
Solving a second-order homogeneous differential equation initial value problem (with distinct real roots)
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Another example with distinct real roots
Example
Find the solution to the differential equation, if ???y(0)=5??? and ???y'(0)=27???.
???3y''+18y'+15y=0???
If we make substitutions for ???y??? in terms of ???r???, we get
???3r^2+18r+15=0???
We’ll factor the left side and solve for ???r???.
???(3r+3)(r+5)=0???
???3r_1+3=0???
???3r_1=-3???
???r_1=-1???
and
???r_2+5=0???
???r_2=-5???
The roots are two real numbers that are unequal (they’re not equal to each other), so these are distinct real roots. Which means we’ll use the formula for the general solution for distinct real roots and get
???y(x)=c_1e^{-x}+c_2e^{-5x}???
This is the general solution to the differential equation, but we still need to use our initial conditions to solve for ???c_1??? and ???c_2???, so we’ll take the derivative of the general solution.
???y'(x)=-c_1e^{-x}-5c_2e^{-5x}???
Applying the initial condition ???y(0)=5??? to the general solution, and the initial condition ???y'(0)=27??? to its derivative, we get
???y(x)=c_1e^{-x}+c_2e^{-5x}???
Since ???y(0)=5???,
???5=c_1e^{-(0)}+c_2e^{-5(0)}???
???5=c_1(1)+c_2(1)???
???5=c_1+c_2???
and
???y'(x)=-c_1e^{-x}-5c_2e^{-5x}???
Since ???y'(0)=27???,
???27=-c_1e^{-(0)}-5c_2e^{-5(0)}???
???27=-c_1(1)-5c_2(1)???
???27=-c_1-5c_2???
We need to solve these as a system of linear equations in order to find values for ???c_1??? and ???c_2???. If we solve ???5=c_1+c_2??? for ???c_1???, we get ???c_1=5-c_2???, and then we can substitute this value into ???27=-c_1-5c_2???.
???27=-c_1-5c_2???
???27=-(5-c_2)-5c_2???
???27=-5+c_2-5c_2???
???32=-4c_2???
???c_2=-8???
Plugging ???c_2=-8??? into ???c_1=5-c_2??? to solve for ???c_1???, we get
???c_1=5-(-8)???
???c_1=13???
Now we can take ???c_1=13??? and ???c_2=-8??? and plug them into the general solution.
???y(x)=c_1e^{-x}+c_2e^{-5x}???
???y(x)=(13)e^{-x}+(-8)e^{-5x}???
???y(x)=13e^{-x}-8e^{-5x}???