Question

In this problem, y = c1ex + c2e−x is a two-parameter family of solutions of the second-order DE y'' − y = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. y(−1) = 9, y'(−1) = −9

Answer 1

Answer:

$y(x)=\frac{9}{e^{1} } e^{-x} =3.310914971e^{-x}$

Step-by-step explanation:

This problem is very simple, since they give the solution for the differential equation from the start. So basically, you need to evaluate the initial conditions into the solution, and the derivative of the solution in order to find the value of the constants $c_1$ and $c_2$.

So, first of all, let's find the derivative of $y(x)$:

$y'(x)=c_1 e^{x} -c_2e^{-x}$

Now, let's evaluate the first initial condition:

$y(-1)=c_1e^{-1} +c_2e^{-(-1)} =9\\\\c_1e^{-1} +c_2e^{1}=9\hspace{10}(1)$

Now, the second initial condition:

$y'(-1)=c_1 e^{-1} -c_2e^{-(-1)}=-9\\\\c_1 e^{-1} -c_2e^{1}=-9\hspace{10}(2)$

Combining (1) and (2) we have a 2x2 System of Equations. Let's use elimination method in order to solve it:

$(1)+(2):\\\\c_1e^{-1} +c_2e^{1} +c_1e^{-1} -c_2e^{1}=9-9\\\\2c_1e^{-1} =0\\\\Hence\\\\c_1=0$

Replacing $c_1$ into (1)

$(0)e^{-1} +c_2e^{1}=9\\\\c_2e^{1}=9\\\\Hence\\\\c_2=\frac{9}{e^{1} } =3.310914971$

Therefore the solution of the second-order IVP is:

$y(x)=\frac{9}{e^{1} } e^{-x} =3.310914971e^{-x}$