Can u proof it pls indeed helpΑΔΒ = ΒΔΑ.
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The system of equations has infinite solutions.
<h2>Given to us,</h2>The value of y is already given in equation 1,
substituting the value of y in equation 2,
The solution of the two equations is 0. Also, we can see that both the equations are in ratio.
Further, the image also shows that the line of the two equations are coinciding.
Hence, the system of equations has infinite solutions.
Learn more about System of solutions:
Answer:
The solution of the differential equation is
Step-by-step explanation:
The differential equation is given by: y" + y = Sin(2t)
<u>i) Using characteristic equation:</u>
The characteristic equation method assumes that y(t)=, where "r" is a constant.
We find the solution of the homogeneus differential equation:
y" + y = 0
As could never be zero, the term (r²+1) must be zero:
(r²+1)=0
r=±i
The solution of the homogeneus differential equation is:
Using Euler's formula:
The particular solution of the differential equation is given by:
So we use these derivatives in the differential equation:
As there is not a term for Cos(2t), B is equal to 0.
So the value A=-1/3
The solution is the sum of the particular function and the homogeneous function:
Using the initial conditions we can check that C1=5/3 and C2=2
<u>ii) Using Laplace Transform:</u>
To solve the differential equation we use the Laplace transformation in both members:
ℒ[y" + y]=ℒ[Sin(2t)]
ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]
By using the Table of Laplace Transform we get:
ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1
ℒ[y]=Y(s)
ℒ[Sin(2t)]=
We replace the previous data in the equation:
s²·Y(s)
-2s-1+Y(s)
=
(s²+1)·Y(s)-2s-1=
(s²+1)·Y(s)=
Y(s)=
Y(s)=
Using partial franction method:
2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)
2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)
We solve the equation system:
A+C=2
B+D=1
A+4C=8
B+4D=6
The solutions are:
A=0 ; B= -2/3 ; C=2 ; D=5/3
So,
Y(s)=
Y(s)=
By using the inverse of the Laplace transform:
ℒ⁻¹[Y(s)]=ℒ⁻¹[]-ℒ⁻¹[
]+ℒ⁻¹[
]