The answer is b because an intersection looks like this:
Hope this helps! <3
Answer:
58.3 ml
Step-by-step explanation:
Step 1: Multiply 362 by 161.
Step 2: Move the decimal point by 3 places to the left.
Step 3: Round to the nearest tenth.
Therefore, the answer is 58.3 ml.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
Answer:
a. because there are no other numbers
Answer:
Therefore the complete primitive is
![y=c_1 e^{2y}+c_2e^{3t}+e^{t}](https://tex.z-dn.net/?f=y%3Dc_1%20e%5E%7B2y%7D%2Bc_2e%5E%7B3t%7D%2Be%5E%7Bt%7D)
Therefore the general solution is
![y=c_1e^{2t}+c_2e^{3t}+e^t](https://tex.z-dn.net/?f=y%3Dc_1e%5E%7B2t%7D%2Bc_2e%5E%7B3t%7D%2Be%5Et)
Step-by-step explanation:
Given Differential equation is
![y''-5y'+6y=2e^t](https://tex.z-dn.net/?f=y%27%27-5y%27%2B6y%3D2e%5Et)
<h3>
Method of variation of parameters:</h3>
Let
be a trial solution.
![y'= me^{mt}](https://tex.z-dn.net/?f=y%27%3D%20me%5E%7Bmt%7D)
and ![y''= m^2e^{mt}](https://tex.z-dn.net/?f=y%27%27%3D%20m%5E2e%5E%7Bmt%7D)
Then the auxiliary equation is
![m^2e^{mt}-5me^{mt}+6e^{mt}=0](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D-5me%5E%7Bmt%7D%2B6e%5E%7Bmt%7D%3D0)
![\Rightarrow m^2-5m+6=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%5E2-5m%2B6%3D0)
![\Rightarrow m^2 -3m -2m +6=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%5E2%20%20-3m%20-2m%20%2B6%3D0)
![\Rightarrow m(m -3) -2(m -3)=0](https://tex.z-dn.net/?f=%5CRightarrow%20m%28m%20%20-3%29%20-2%28m%20-3%29%3D0)
![\Rightarrow (m-3)(m-2)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%20%28m-3%29%28m-2%29%3D0)
![\Rightarrow m=2,3](https://tex.z-dn.net/?f=%5CRightarrow%20%20m%3D2%2C3)
∴The complementary function is ![C_1e^{2t}+C_2e^{3t}](https://tex.z-dn.net/?f=C_1e%5E%7B2t%7D%2BC_2e%5E%7B3t%7D)
To find P.I
First we show that
and
are linearly independent solution.
Let
and ![y_2= e^{3t}](https://tex.z-dn.net/?f=y_2%3D%20e%5E%7B3t%7D)
The Wronskian of
and
is ![\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bcc%7Dy_1%26y_2%5C%5Cy%27_1%26y%27_2%5Cend%7Barray%7D%5Cright%7C)
![=\left|\begin{array}{cc}e^{2t}&e^{3t}\\2e^{2t}&3e^{3t}\end{array}\right|](https://tex.z-dn.net/?f=%3D%5Cleft%7C%5Cbegin%7Barray%7D%7Bcc%7De%5E%7B2t%7D%26e%5E%7B3t%7D%5C%5C2e%5E%7B2t%7D%263e%5E%7B3t%7D%5Cend%7Barray%7D%5Cright%7C)
![=e^{2t}.3e^{3t}-e^{2t}.2e^{3t}](https://tex.z-dn.net/?f=%3De%5E%7B2t%7D.3e%5E%7B3t%7D-e%5E%7B2t%7D.2e%5E%7B3t%7D)
≠ 0
∴
and
are linearly independent.
Let the particular solution is
![y_p=v_1(t)e^{2t}+v_2(t)e^{3t}](https://tex.z-dn.net/?f=y_p%3Dv_1%28t%29e%5E%7B2t%7D%2Bv_2%28t%29e%5E%7B3t%7D)
Then,
![Dy_p= 2v_1(t)e^{2t}+v'_1(t)e^{2t}+3v_2(t)e^{3t}+v'_2(t)e^{3t}](https://tex.z-dn.net/?f=Dy_p%3D%202v_1%28t%29e%5E%7B2t%7D%2Bv%27_1%28t%29e%5E%7B2t%7D%2B3v_2%28t%29e%5E%7B3t%7D%2Bv%27_2%28t%29e%5E%7B3t%7D)
Choose
and
such that
.......(1)
So that
![Dy_p= 2v_1(t)e^{2t}+3v_2(t)e^{3t}](https://tex.z-dn.net/?f=Dy_p%3D%202v_1%28t%29e%5E%7B2t%7D%2B3v_2%28t%29e%5E%7B3t%7D)
![D^2y_p= 4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}](https://tex.z-dn.net/?f=D%5E2y_p%3D%204v_1%28t%29e%5E%7B2t%7D%2B9v_2%28t%29e%5E%7B3t%7D%2B%202v%27_1%28t%29e%5E%7B2t%7D%2B3v%27_2%28t%29e%5E%7B3t%7D)
Now
![4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}-5[2v_1(t)e^{2t}+3v_2(t)e^{3t}] +6[v_1e^{2t}+v_2e^{3t}]=2e^t](https://tex.z-dn.net/?f=4v_1%28t%29e%5E%7B2t%7D%2B9v_2%28t%29e%5E%7B3t%7D%2B%202v%27_1%28t%29e%5E%7B2t%7D%2B3v%27_2%28t%29e%5E%7B3t%7D-5%5B2v_1%28t%29e%5E%7B2t%7D%2B3v_2%28t%29e%5E%7B3t%7D%5D%20%2B6%5Bv_1e%5E%7B2t%7D%2Bv_2e%5E%7B3t%7D%5D%3D2e%5Et)
.......(2)
Solving (1) and (2) we get
and ![v'_1(t)=-2e^{-t}](https://tex.z-dn.net/?f=v%27_1%28t%29%3D-2e%5E%7B-t%7D)
Hence
![v_1(t)=\int (-2e^{-t}) dt=2e^{-t}](https://tex.z-dn.net/?f=v_1%28t%29%3D%5Cint%20%28-2e%5E%7B-t%7D%29%20dt%3D2e%5E%7B-t%7D)
and ![v_2=\int 2e^{-2t}dt =-e^{-2t}](https://tex.z-dn.net/?f=v_2%3D%5Cint%202e%5E%7B-2t%7Ddt%20%3D-e%5E%7B-2t%7D)
Therefore ![y_p=(2e^{-t}) e^{2t}-e^{-2t}.e^{3t}](https://tex.z-dn.net/?f=y_p%3D%282e%5E%7B-t%7D%29%20e%5E%7B2t%7D-e%5E%7B-2t%7D.e%5E%7B3t%7D)
![=2e^t-e^t](https://tex.z-dn.net/?f=%3D2e%5Et-e%5Et)
![=e^t](https://tex.z-dn.net/?f=%3De%5Et)
Therefore the complete primitive is
![y=c_1 e^{2y}+c_2e^{3t}+ e^{t}](https://tex.z-dn.net/?f=y%3Dc_1%20e%5E%7B2y%7D%2Bc_2e%5E%7B3t%7D%2B%20e%5E%7Bt%7D)
<h3>
Undermined coefficients:</h3>
∴The complementary function is ![C_1e^{2t}+C_2e^{3t}](https://tex.z-dn.net/?f=C_1e%5E%7B2t%7D%2BC_2e%5E%7B3t%7D)
The particular solution is ![y_p=Ae^t](https://tex.z-dn.net/?f=y_p%3DAe%5Et)
Then,
and ![D^2y_p=Ae^t](https://tex.z-dn.net/?f=D%5E2y_p%3DAe%5Et)
![\therefore Ae^t-5Ae^t+6Ae^t=2e^t](https://tex.z-dn.net/?f=%5Ctherefore%20Ae%5Et-5Ae%5Et%2B6Ae%5Et%3D2e%5Et)
![\Rightarrow 2Ae^t=2e^t](https://tex.z-dn.net/?f=%5CRightarrow%202Ae%5Et%3D2e%5Et)
![\therefore y_p=e^t](https://tex.z-dn.net/?f=%5Ctherefore%20y_p%3De%5Et)
Therefore the general solution is
![y=c_1e^{2t}+c_2e^{3t}+e^t](https://tex.z-dn.net/?f=y%3Dc_1e%5E%7B2t%7D%2Bc_2e%5E%7B3t%7D%2Be%5Et)