Since derivatives lower the degree of polynomials, if you want the second derivative to be zero you have to choose first-degree polynomials.
So, you have
![y''(x)=0 \iff y = c_1x+c_0](https://tex.z-dn.net/?f=y%27%27%28x%29%3D0%20%5Ciff%20y%20%3D%20c_1x%2Bc_0)
Two polynomials are linearly independent if they are not multiples of each other. So, for example, you might choose
and
to find two linearly independent solutions.
As for
![y''(x)=1](https://tex.z-dn.net/?f=y%27%27%28x%29%3D1)
we want a second-degree polynomial with leading coefficient 1/2 so that we will get 1 when deriving it twice:
![y''(x)=1 \iff y(x)=\dfrac{x^2}{2}+c_1x+c_0](https://tex.z-dn.net/?f=y%27%27%28x%29%3D1%20%5Ciff%20y%28x%29%3D%5Cdfrac%7Bx%5E2%7D%7B2%7D%2Bc_1x%2Bc_0)
If we impose the conditions
![y(0)=0,\quad y'(0)=0](https://tex.z-dn.net/?f=y%280%29%3D0%2C%5Cquad%20y%27%280%29%3D0)
we have
![y(0)=\dfrac{0^2}{2}+c_1\cdot 0+c_0 = c_0 = 0](https://tex.z-dn.net/?f=y%280%29%3D%5Cdfrac%7B0%5E2%7D%7B2%7D%2Bc_1%5Ccdot%200%2Bc_0%20%3D%20c_0%20%3D%200)
So, our solution will be in this form:
![y''(x)=1,\quad y(0)=0 \iff y(x)=\dfrac{x^2}{2}+c_1x](https://tex.z-dn.net/?f=y%27%27%28x%29%3D1%2C%5Cquad%20y%280%29%3D0%20%5Ciff%20y%28x%29%3D%5Cdfrac%7Bx%5E2%7D%7B2%7D%2Bc_1x)
To fix
, we use the second condition:
![y'(x) = x+c_1 \implies y'(0) = c_1 = 0](https://tex.z-dn.net/?f=y%27%28x%29%20%3D%20x%2Bc_1%20%5Cimplies%20y%27%280%29%20%3D%20c_1%20%3D%200)
So, we have fixed
and the solutions is
![\begin{cases}y''(x)=1\\y'(0)=0\\y(0)=0\end{cases} \iff y(x) = \dfrac{x^2}{2}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dy%27%27%28x%29%3D1%5C%5Cy%27%280%29%3D0%5C%5Cy%280%29%3D0%5Cend%7Bcases%7D%20%5Ciff%20y%28x%29%20%3D%20%5Cdfrac%7Bx%5E2%7D%7B2%7D)
Answer:20%
Step-by-step explanation:
You add 72 plus 18 and get 100 then you find what 100 divided by 5 is.
Probability of having a club:
1st card = 13/52
2nd card= 12/51
3rd card = 11/50
Total conditional probability = 13/52 * 12/51 * 11/50= 0.0129 =1.29%
Answer:
i think tat the answer is C