Answer:
The complete solution is
Step-by-step explanation:
Given differential equation is
3y"- 8y' - 3y =4
The trial solution is
![y = e^{mx}](https://tex.z-dn.net/?f=y%20%3D%20e%5E%7Bmx%7D)
Differentiating with respect to x
![y'= me^{mx}](https://tex.z-dn.net/?f=y%27%3D%20me%5E%7Bmx%7D)
Again differentiating with respect to x
![y''= m ^2 e^{mx}](https://tex.z-dn.net/?f=y%27%27%3D%20m%20%5E2%20e%5E%7Bmx%7D)
Putting the value of y, y' and y'' in left side of the differential equation
![3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0](https://tex.z-dn.net/?f=3m%5E2e%5E%7Bmx%7D-8m%20e%5E%7Bmx%7D-%203e%5E%7Bmx%7D%3D0)
![\Rightarrow 3m^2-8m-3=0](https://tex.z-dn.net/?f=%5CRightarrow%203m%5E2-8m-3%3D0)
The auxiliary equation is
![3m^2-8m-3=0](https://tex.z-dn.net/?f=3m%5E2-8m-3%3D0)
![\Rightarrow 3m^2 -9m+m-3m=0](https://tex.z-dn.net/?f=%5CRightarrow%203m%5E2%20-9m%2Bm-3m%3D0)
![\Rightarrow 3m(m-3)+1(m-3)=0](https://tex.z-dn.net/?f=%5CRightarrow%203m%28m-3%29%2B1%28m-3%29%3D0)
![\Rightarrow (3m+1)(m-3)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%283m%2B1%29%28m-3%29%3D0)
The complementary function is
![y= Ae^{3x}+Be^{-\frac13 x}](https://tex.z-dn.net/?f=y%3D%20Ae%5E%7B3x%7D%2BBe%5E%7B-%5Cfrac13%20x%7D)
y''= D², y' = D
The given differential equation is
(3D²-8D-3D)y =4
⇒(3D+1)(D-3)y =4
Since the linear operation is
L(D) ≡ (3D+1)(D-3)
For particular integral
![y_p=\frac 1{(3D+1)(D-3)} .4](https://tex.z-dn.net/?f=y_p%3D%5Cfrac%201%7B%283D%2B1%29%28D-3%29%7D%20.4)
[since
]
[ replace D by 0 , since L(0)≠0]
![=-\frac43](https://tex.z-dn.net/?f=%3D-%5Cfrac43)
The complete solution is
y= C.F+P.I
![\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43](https://tex.z-dn.net/?f=%5Ctherefore%20y%3D%20Ae%5E%7B3x%7D%2BBe%5E%7B-%5Cfrac13%20x%7D-%5Cfrac43)
Answer:
11/8 OR 1 3/8
Step-by-step explanation:
Given equations are;<span>
2a + 3b = -1 ..................equation 1
3a + 5b = -2 ....................equation 2</span>
Now multiply equation 1 with (-3)
The equation will be;
-6a -9b = 3 …………………..equation 3
Now multiply equation 2 with (2)
The equation will be;
6a + 10b = -4 ……………..equation 4
Now add equation 3 and equation 4
-6a – 9b = 3
<span>6a + 10b = -4</span>
<span>------------------------------</span>
0a + b = -1
b = -1
Now put the value of b in equation 1
2a + 3(-1) = -1
2a -3 = -1
2a = -1+3
2a = 2
a=1
Thus the solution is (a,b) = (1,-1)
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Answer:
sup
Step-by-step explanation: