The angle would be 50 degrees
Answer:
satisfies given homogenous solution and the particular solution is
.
Step-by-step explanation:
Given homogeneous equation is,
where t>0 subject to ![y_1(t)=t^2, y_2(t)=\frac{1}{t}\hfill (1)](https://tex.z-dn.net/?f=y_1%28t%29%3Dt%5E2%2C%20y_2%28t%29%3D%5Cfrac%7B1%7D%7Bt%7D%5Chfill%20%281%29)
To verify,
- wheather
satisfies given homogenous equation, then both will satisfy
that is,
![t^2y_{1}^{''}-2y_1=t^2\times 2-2\times t^2=0](https://tex.z-dn.net/?f=t%5E2y_%7B1%7D%5E%7B%27%27%7D-2y_1%3Dt%5E2%5Ctimes%202-2%5Ctimes%20t%5E2%3D0)
![t^2y_{2}^{''}-2y_2=t^2\times 2t^{-3}-2\times t^{-1}=2t^{-1}-2t^{-1}=0](https://tex.z-dn.net/?f=t%5E2y_%7B2%7D%5E%7B%27%27%7D-2y_2%3Dt%5E2%5Ctimes%202t%5E%7B-3%7D-2%5Ctimes%20t%5E%7B-1%7D%3D2t%5E%7B-1%7D-2t%5E%7B-1%7D%3D0)
Thus
and
satisfies (1).
Now the wronskean,
![W(y_1, y_2)(x)=y_1y_{2}^{'}-y_{2}y_{1}^{'}=-t^2t^{-2}-2tt^{-1}=-3\neq 0](https://tex.z-dn.net/?f=W%28y_1%2C%20y_2%29%28x%29%3Dy_1y_%7B2%7D%5E%7B%27%7D-y_%7B2%7Dy_%7B1%7D%5E%7B%27%7D%3D-t%5E2t%5E%7B-2%7D-2tt%5E%7B-1%7D%3D-3%5Cneq%200)
Thus
and
are solution of (1) .
where ![D\equiv \frac{\partial }{\partial t}](https://tex.z-dn.net/?f=D%5Cequiv%20%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20t%7D)
![=-\frac{1}{2t^2}\frac{1}{1-\frac{D^2}{t^2}}()5t^2-1](https://tex.z-dn.net/?f=%3D-%5Cfrac%7B1%7D%7B2t%5E2%7D%5Cfrac%7B1%7D%7B1-%5Cfrac%7BD%5E2%7D%7Bt%5E2%7D%7D%28%295t%5E2-1)
![=-\frac{1}{2t^2}(1-\frac{D^2}{t^2})^{-1}(5t^2-1)](https://tex.z-dn.net/?f=%3D-%5Cfrac%7B1%7D%7B2t%5E2%7D%281-%5Cfrac%7BD%5E2%7D%7Bt%5E2%7D%29%5E%7B-1%7D%285t%5E2-1%29)
![=-\frac{1}{2t^2}(1-\frac{D^2}{t^2}+........)(5t^2-1)](https://tex.z-dn.net/?f=%3D-%5Cfrac%7B1%7D%7B2t%5E2%7D%281-%5Cfrac%7BD%5E2%7D%7Bt%5E2%7D%2B........%29%285t%5E2-1%29)
sincce ![D^2(5t^2-1)=10](https://tex.z-dn.net/?f=D%5E2%285t%5E2-1%29%3D10)
![=\frac{5}{t^4}+\frac{1}{2t^2}-\frac{5}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B5%7D%7Bt%5E4%7D%2B%5Cfrac%7B1%7D%7B2t%5E2%7D-%5Cfrac%7B5%7D%7B2%7D)
Hence the result.
![5\cdot(\frac{1}{5^3})\ne5\cdot(5^3)](https://tex.z-dn.net/?f=5%5Ccdot%28%5Cfrac%7B1%7D%7B5%5E3%7D%29%5Cne5%5Ccdot%285%5E3%29)
His answer is not correct, when we have a power on the denominator it is dividing and not multiplying, in this case while the expression on the left is being divided the other one is multiplying.
also you could rewrite the expressions like so remembering that powers that are on the denominator can be written as negative powers.
Answer:
30% of 80
Step-by-step explanation:
Answer:
16 i think
Step-by-step explanation: