Step-by-step explanation:
(x+5)^2 = 2( 5x-3)
x^2+25=10x-6
x^2-10x+25+6= 0
x^2-10x+31=0
a=1, b
Answer:
5.15
Step-by-step explanation:
17.51 / 3.4 = 5.15
For the ODE
![ty'+2y=\sin t](https://tex.z-dn.net/?f=ty%27%2B2y%3D%5Csin%20t)
multiply both sides by <em>t</em> so that the left side can be condensed into the derivative of a product:
![t^2y'+2ty=t\sin t](https://tex.z-dn.net/?f=t%5E2y%27%2B2ty%3Dt%5Csin%20t)
![\implies(t^2y)'=t\sin t](https://tex.z-dn.net/?f=%5Cimplies%28t%5E2y%29%27%3Dt%5Csin%20t)
Integrate both sides with respect to <em>t</em> :
![t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C](https://tex.z-dn.net/?f=t%5E2y%3D%5Cdisplaystyle%5Cint%20t%5Csin%20t%5C%2C%5Cmathrm%20dt%3D%5Csin%20t-t%5Ccos%20t%2BC)
Divide both sides by
to solve for <em>y</em> :
![y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B%5Csin%20t%7D%7Bt%5E2%7D-%5Cdfrac%7B%5Ccos%20t%7Dt%2B%5Cdfrac%20C%7Bt%5E2%7D)
Now use the initial condition to solve for <em>C</em> :
![y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}](https://tex.z-dn.net/?f=y%5Cleft%28%5Cdfrac%5Cpi2%5Cright%29%3D9%5Cimplies9%3D%5Cdfrac%7B%5Csin%5Cfrac%5Cpi2%7D%7B%5Cfrac%7B%5Cpi%5E2%7D4%7D-%5Cdfrac%7B%5Ccos%5Cfrac%5Cpi2%7D%7B%5Cfrac%5Cpi2%7D%2B%5Cdfrac%20C%7B%5Cfrac%7B%5Cpi%5E2%7D4%7D)
![\implies9=\dfrac4{\pi^2}(1+C)](https://tex.z-dn.net/?f=%5Cimplies9%3D%5Cdfrac4%7B%5Cpi%5E2%7D%281%2BC%29)
![\implies C=\dfrac{9\pi^2}4-1](https://tex.z-dn.net/?f=%5Cimplies%20C%3D%5Cdfrac%7B9%5Cpi%5E2%7D4-1)
So the particular solution to the IVP is
![y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B%5Csin%20t%7D%7Bt%5E2%7D-%5Cdfrac%7B%5Ccos%20t%7Dt%2B%5Cdfrac%7B%5Cfrac%7B9%5Cpi%5E2%7D4-1%7D%7Bt%5E2%7D)
or
![y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cdfrac%7B4%5Csin%20t-4t%5Ccos%20t%2B9%5Cpi%5E2-4%7D%7B4t%5E2%7D)