Find the general solution of the given differential equations 4y′′-4y′-3y=0

William Curry 2021-12-29 Answered
Find the general solution of the given differential equations
4y4y3y=0
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Expert Answer

Kayla Kline
Answered 2021-12-30 Author has 37 answers
Step 1
Given differential equation is
4y4y3y=0
Step 2
Auxiliary equation of differential equation is
4m24m3=0
4m2+2m6m3=0
2m(2m+1)3(2m+1)=0
(2m+1)(2m3)=0
m=12,32
Hence general solution of differential equation is
y=C1e32t+C2e12t
Step 3
Answer:
General solution of given differential equation is y=C1e32t+C2e12t

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usumbiix
Answered 2021-12-31 Author has 33 answers
Solving the characteristic equation and applying the initial conditions, the solution of the IVP is: y(t)=1.75e0.5t+2.75e1.5t
The IVP is given by:
4y4y3y=0
The characteristic equation is:
4r24r3=0
Which is a quadratic equation with coefficients a=4,b=4,c=3, and thus:
=b24ac=(4)24(4)(3)=64
x1=b2a=488=0.5
x2=b2a=4+88=1.5
Thus, the solution is in the format:
y(t)=ae0.5t+be1.5t
Applying the initial conditions:
y(0)=1a+b=1
y(0)=50.5a+1.5b=5
From the first equation, a=1b, and replacing in the second:
0.5a+1.5b=5
0.5+0.5b+1.5b=5
2b=5.5
b=5.52
b=2.75
a=1b=12.75=1.75
Then, the solution to the IVP is:
y(t)=1.75e0.5t+2.75e1.5t

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karton
Answered 2022-01-09 Author has 439 answers

Step-by-step explanation:
Given:
4y4y3y=0,y(0)=1,y(0)=5
Now, the auxiliary equation will be given as:
4m24m3=0
or 4m2+2m6m3=0
or 2m(2+1)3(2m+1)=0
or (2m3)×(2m+1)=0
therefore, m=32 and m=12
thus, the general equation comes as: y=C1emx+C2emx
or y=C1e12x+C2e32x
Now, at y(0)=1
Therefore, 1=C1e12×0+C2e32×0
or C1+C2=1(1)
and, y=12C1e12x+32C2e32x
Also, y(0)=5
Thus, 5=12C1e12×0+32C2e32×0
or 3C2C1=10(2)
on adding 1 and 2, we get C1+C2=1
+(C1+3C2)=10
4C2=11
or C2=114
Thus, C1+C2=1
or C1+114=1
or C1=74
Hence, the solution is y=C1emx+C2emx
on substituting the values,
y=74e12x+114e32x

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