To take advantage of the characteristic solutions
and
, you can try the method of variation of parameters, where we look for a solution of the form
![y=y_1u_1+y_2u_2](https://tex.z-dn.net/?f=y%3Dy_1u_1%2By_2u_2)
with the condition that
![(\mathbf 1)](https://tex.z-dn.net/?f=%28%5Cmathbf%201%29)
Then
![y'={y_1}'u_1+y_1{u_1}'+{y_2}'u_2+y_2{u_2}'](https://tex.z-dn.net/?f=y%27%3D%7By_1%7D%27u_1%2By_1%7Bu_1%7D%27%2B%7By_2%7D%27u_2%2By_2%7Bu_2%7D%27)
![\implies y'={y_1}'u_1+{y_2}'u_2](https://tex.z-dn.net/?f=%5Cimplies%20y%27%3D%7By_1%7D%27u_1%2B%7By_2%7D%27u_2)
![y''={y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}'](https://tex.z-dn.net/?f=y%27%27%3D%7By_1%7D%27%27u_1%2B%7By_1%7D%27%7Bu_1%7D%27%2B%7By_2%7D%27%27u_2%2B%7By_2%7D%27%7Bu_2%7D%27)
Substituting into the ODE gives
![(1-x)({y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}')+x({y_1}'u_1+{y_2}'u_2)-y_1u_1+y_2u_2=2(x-1)^2e^{-x}](https://tex.z-dn.net/?f=%281-x%29%28%7By_1%7D%27%27u_1%2B%7By_1%7D%27%7Bu_1%7D%27%2B%7By_2%7D%27%27u_2%2B%7By_2%7D%27%7Bu_2%7D%27%29%2Bx%28%7By_1%7D%27u_1%2B%7By_2%7D%27u_2%29-y_1u_1%2By_2u_2%3D2%28x-1%29%5E2e%5E%7B-x%7D)
Since
![y_1=x\implies{y_1}'=1\implies{y_1}''=0](https://tex.z-dn.net/?f=y_1%3Dx%5Cimplies%7By_1%7D%27%3D1%5Cimplies%7By_1%7D%27%27%3D0)
![y_2=e^x\implies{y_2}'=e^x\implies{y_2}''=e^x](https://tex.z-dn.net/?f=y_2%3De%5Ex%5Cimplies%7By_2%7D%27%3De%5Ex%5Cimplies%7By_2%7D%27%27%3De%5Ex)
the above reduces to
![(1-x)({u_1}'+e^x{u_2}')=2(x-1)^2e^{-x}](https://tex.z-dn.net/?f=%281-x%29%28%7Bu_1%7D%27%2Be%5Ex%7Bu_2%7D%27%29%3D2%28x-1%29%5E2e%5E%7B-x%7D)
![(\mathbf 2)](https://tex.z-dn.net/?f=%28%5Cmathbf%202%29)
and
form a linear system that we can solve for
using Cramer's rule:
![{u_1}'=\dfrac{W_1(x)}{W(x)},{u_2}'=\dfrac{W_2(x)}{W(x)}](https://tex.z-dn.net/?f=%7Bu_1%7D%27%3D%5Cdfrac%7BW_1%28x%29%7D%7BW%28x%29%7D%2C%7Bu_2%7D%27%3D%5Cdfrac%7BW_2%28x%29%7D%7BW%28x%29%7D)
where
is the Wronskian determinant of the fundamental system and
is the same determinant, but with the
-th column replaced with
.
![W(x)=\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=e^x(x-1)](https://tex.z-dn.net/?f=W%28x%29%3D%5Cbegin%7Bvmatrix%7Dx%26e%5Ex%5C%5C1%26e%5Ex%5Cend%7Bvmatrix%7D%3De%5Ex%28x-1%29)
![W_1(x)=\begin{vmatrix}0&e^x\\2(x-1)^2e^{-x}&e^x\end{vmatrix}=-2(x-1)^2](https://tex.z-dn.net/?f=W_1%28x%29%3D%5Cbegin%7Bvmatrix%7D0%26e%5Ex%5C%5C2%28x-1%29%5E2e%5E%7B-x%7D%26e%5Ex%5Cend%7Bvmatrix%7D%3D-2%28x-1%29%5E2)
![W_2(x)=\begin{vmatrix}x&0\\e^x&2(x-1)^2e^{-x}\end{vmatrix}=2xe^{-x}(x-1)^2](https://tex.z-dn.net/?f=W_2%28x%29%3D%5Cbegin%7Bvmatrix%7Dx%260%5C%5Ce%5Ex%262%28x-1%29%5E2e%5E%7B-x%7D%5Cend%7Bvmatrix%7D%3D2xe%5E%7B-x%7D%28x-1%29%5E2)
So we have
![{u_1}'=\dfrac{-2(x-1)^2}{e^x(x-1)}\implies u_1=2xe^{-x}](https://tex.z-dn.net/?f=%7Bu_1%7D%27%3D%5Cdfrac%7B-2%28x-1%29%5E2%7D%7Be%5Ex%28x-1%29%7D%5Cimplies%20u_1%3D2xe%5E%7B-x%7D)
![{u_2}'=\dfrac{2xe^{-x}(x-1)^2}{e^x(x-1)}\implies u_2=-x^2e^{-2x}](https://tex.z-dn.net/?f=%7Bu_2%7D%27%3D%5Cdfrac%7B2xe%5E%7B-x%7D%28x-1%29%5E2%7D%7Be%5Ex%28x-1%29%7D%5Cimplies%20u_2%3D-x%5E2e%5E%7B-2x%7D)
Then the particular solution is
![y_p=2x^2e^{-x}-x^2e^{-x}=x^2e^{-x}](https://tex.z-dn.net/?f=y_p%3D2x%5E2e%5E%7B-x%7D-x%5E2e%5E%7B-x%7D%3Dx%5E2e%5E%7B-x%7D)
giving the general solution to the ODE,
![\boxed{y(x)=C_1x+C_2e^x+x^2e^{-x}}](https://tex.z-dn.net/?f=%5Cboxed%7By%28x%29%3DC_1x%2BC_2e%5Ex%2Bx%5E2e%5E%7B-x%7D%7D)
Answer:A = 1500e^(0.03 * 5) = 1500e^(0.15)
use definition of logarithm
0.15 = ln(A / 1500)
0.15 = ln(A) - ln(1500)
ln(A) = 0.15 + ln(1500) = 0.15 + 7.313220387 = 7.463220387
Hi Dez123,
Graph the system of equations. {−x+2y=−8 3x−y=−6
−y= −3x + 8
6x − 2y = 8
Multiply y = 3x − 8 by -1
y = 3x − 8
6x − 2y = 8
y = 3x − 8
−2y = −6x + 8
y = 3x − 8
y = 3x − 4
The graph is in the picture below!
Hope This Helps!