The sum total of the angles in a triangle is equal to 180° degrees.
Answer:
Monica spent 0.55 hours listening to Brahms.
Step-by-step explanation:
We are given the following in the question:
Amount of time spent listening to tapes of Beethoven and Brahms =
![=\dfrac{3}{4}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B3%7D%7B4%7D)
Amount of time spent listening to Beethoven =
![=\dfrac{1}{5}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1%7D%7B5%7D)
Total time spent listening to Brahms =
Amount of time spent listening to tapes of Beethoven and Brahms - Amount of time spent listening to Beethoven
![=\dfrac{3}{4}-\dfrac{1}{5}\\\\=\dfrac{15-4}{20}\\\\=\dfrac{11}{20}\text{ Hours}\\\\=0.55\text{ Hours}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B3%7D%7B4%7D-%5Cdfrac%7B1%7D%7B5%7D%5C%5C%5C%5C%3D%5Cdfrac%7B15-4%7D%7B20%7D%5C%5C%5C%5C%3D%5Cdfrac%7B11%7D%7B20%7D%5Ctext%7B%20Hours%7D%5C%5C%5C%5C%3D0.55%5Ctext%7B%20Hours%7D)
Thus, Monica spent
listening to Brahms.
The corresponding homogeneous ODE has characteristic equation
with roots at
, thus admitting the characteristic solution
![y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x](https://tex.z-dn.net/?f=y_c%3DC_1e%5Ex%5Ccos%5Cdfrac%7B%5Csqrt3%7D2x%2BC_2e%5Ex%5Csin%5Cdfrac%7B%5Csqrt3%7D2x)
For the particular solution, assume one of the form
![y_p=a\sin x+b\cos x](https://tex.z-dn.net/?f=y_p%3Da%5Csin%20x%2Bb%5Ccos%20x)
![{y_p}'=a\cos x-b\sin x](https://tex.z-dn.net/?f=%7By_p%7D%27%3Da%5Ccos%20x-b%5Csin%20x)
![{y_p}''=-a\sin x-b\cos x](https://tex.z-dn.net/?f=%7By_p%7D%27%27%3D-a%5Csin%20x-b%5Ccos%20x)
Substituting into the ODE gives
![(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x](https://tex.z-dn.net/?f=%28-a%5Csin%20x-b%5Ccos%20x%29-%28a%5Ccos%20x-b%5Csin%20x%29%2B%28a%5Csin%20x%2Bb%5Ccos%20x%29%3D%5Csin%20x)
![-b\cos x+a\sin x=\sin x](https://tex.z-dn.net/?f=-b%5Ccos%20x%2Ba%5Csin%20x%3D%5Csin%20x)
![\implies a=1,b=0](https://tex.z-dn.net/?f=%5Cimplies%20a%3D1%2Cb%3D0)
Then the general solution to this ODE is
![\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}](https://tex.z-dn.net/?f=%5Cboxed%7By%28x%29%3DC_1e%5Ex%5Ccos%5Cdfrac%7B%5Csqrt3%7D2x%2BC_2e%5Ex%5Csin%5Cdfrac%7B%5Csqrt3%7D2x%2B%5Csin%20x%7D)
![\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2](https://tex.z-dn.net/?f=%5Cimplies%20r%5E2-3r%2B2%3D%28r-1%29%28r-2%29%3D0%5Cimplies%20r%3D1%2Cr%3D2)
![\implies y_c=C_1e^x+C_2e^{2x}](https://tex.z-dn.net/?f=%5Cimplies%20y_c%3DC_1e%5Ex%2BC_2e%5E%7B2x%7D)
Assume a solution of the form
![y_p=e^x(a\sin x+b\cos x)](https://tex.z-dn.net/?f=y_p%3De%5Ex%28a%5Csin%20x%2Bb%5Ccos%20x%29)
![{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)](https://tex.z-dn.net/?f=%7By_p%7D%27%3De%5Ex%28%28a%2Bb%29%5Ccos%20x%2B%28a-b%29%5Csin%20x%29)
![{y_p}''=2e^x(a\cos x-b\sin x)](https://tex.z-dn.net/?f=%7By_p%7D%27%27%3D2e%5Ex%28a%5Ccos%20x-b%5Csin%20x%29)
Substituting into the ODE gives
![2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x](https://tex.z-dn.net/?f=2e%5Ex%28a%5Ccos%20x-b%5Csin%20x%29-3e%5Ex%28%28a%2Bb%29%5Ccos%20x%2B%28a-b%29%5Csin%20x%29%2B2e%5Ex%28a%5Csin%20x%2Bb%5Ccos%20x%29%3De%5Ex%5Csin%20x)
![-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x](https://tex.z-dn.net/?f=-e%5Ex%28%28a%2Bb%29%5Ccos%20x%2B%28a-b%29%5Csin%20x%29%3De%5Ex%5Csin%20x)
![\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12](https://tex.z-dn.net/?f=%5Cimplies%5Cbegin%7Bcases%7D-a-b%3D0%5C%5C-a%2Bb%3D1%5Cend%7Bcases%7D%5Cimplies%20a%3D-%5Cdfrac12%2Cb%3D%5Cdfrac12)
so the solution is
![\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}](https://tex.z-dn.net/?f=%5Cboxed%7By%28x%29%3DC_1e%5Ex%2BC_2e%5E%7B2x%7D-%5Cdfrac%7Be%5Ex%7D2%28%5Csin%20x-%5Ccos%20x%29%7D)
![r^2+1=0\implies r=\pm i](https://tex.z-dn.net/?f=r%5E2%2B1%3D0%5Cimplies%20r%3D%5Cpm%20i)
![\implies y_c=C_1\cos x+C_2\sin x](https://tex.z-dn.net/?f=%5Cimplies%20y_c%3DC_1%5Ccos%20x%2BC_2%5Csin%20x)
Assume a solution of the form
![y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)](https://tex.z-dn.net/?f=y_p%3D%28ax%2Bb%29%5Ccos%282x%29%2B%28cx%2Bd%29%5Csin%282x%29)
![{y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)](https://tex.z-dn.net/?f=%7By_p%7D%27%27%3D-4%28ax%2Bb-c%29%5Ccos%282x%29-4%28cx%2Ba%2Bd%29%5Csin%282x%29)
Substituting into the ODE gives
![(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)](https://tex.z-dn.net/?f=%28-4%28ax%2Bb-c%29%5Ccos%282x%29-4%28cx%2Ba%2Bd%29%5Csin%282x%29%29%2B%28%28ax%2Bb%29%5Ccos%282x%29%2B%28cx%2Bd%29%5Csin%282x%29%29%3Dx%5Ccos%282x%29)
![-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)](https://tex.z-dn.net/?f=-%283ax%2B3b-4c%29%5Ccos%282x%29-%283cx%2B3d%2B4a%29%5Csin%282x%29%3Dx%5Ccos%282x%29)
![\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49](https://tex.z-dn.net/?f=%5Cimplies%5Cbegin%7Bcases%7D-3a%3D1%5C%5C-3b%2B4c%3D0%5C%5C-3c%3D0%5C%5C-4a-3d%3D0%5Cend%7Bcases%7D%5Cimplies%20a%3D-%5Cdfrac13%2Cb%3Dc%3D0%2Cd%3D%5Cdfrac49)
so the solution is
![\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}](https://tex.z-dn.net/?f=%5Cboxed%7By%28x%29%3DC_1%5Ccos%20x%2BC_2%5Csin%20x-%5Cdfrac13x%5Ccos%282x%29%2B%5Cdfrac49%5Csin%282x%29%7D)
Answer:
m=-3
Step-by-step explanation:
Since the lines are parallel to each other, their gradients are equal.
∴ gradient of green line = gradient of red line
∴ m = -3