Find the complete solution to the 2nd ODE:
2y''+3y'+y=t^(2)+3sin(t)
Start with solving for the complementary solution to the homogeneous equation
2y"+3y'+y=0,
the characteristic equation is
2r²+3r+1=0
which is an algebraic equation which can be factorized into
(2r+1)(r+1)=0
or
r={-(1/2), -1}
The complementary solution is therefore:
yc=Ae-(x/2) + Be-x
The next step is to find the particular solution for the right-hand side:
y²+3sin(t)
For this, we will need to assume that the particular solution yp is as follows:
yp=Ct²+Dt+Esin(t)+Fcos(t)+G
Now to find the constants C,D,E,F,G, we need to calculate
2yp"+3yp'+yp=t²+sin(t)
and equate coefficients.
We end up with:
G+(-3sin(t)-cos(t))F+(3cos(t)-sin(t))E+(t+3)D+(t^2+6*t+4)C = t²+sin(t)
which can solved for C,D,E,F,G by comparing coefficients and successive substitution:
Sum coefficients of sin(t)=1,
sum of coefficients of cos(t)=0:
-3F-E=1
-F+3E=0
which when solved, gives
E=-1/10, F=-3/10.
Now tackle C, D and G:
Compare coefficients of t² yields
C=1
Substitute C and compare coefficients of t gives D=-6
and consequently coefficient G=14.
The general solution is therefore:
y(t)=yc+yp
=Ae-(t/2) + Be-t +
t²-6t -sin(t)/10 -3cos(t)/10+14
Finally, we check that
2y"(t)+3y'(t)+y(t) equals t²+sin(t)
which confirms that the solution is correct.
If you need help with finding the standard forms of the particular solution, check your course notes, your textbook, or the following link:
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
To find the complete solution to the given second-order linear homogeneous differential equation (ODE), we need to follow these steps:
1. Find the general solution to the associated homogeneous equation, which is obtained by setting the right-hand side equal to zero: 2y'' + 3y' + y = 0.
2. Find one particular solution to the non-homogeneous equation: 2y'' + 3y' + y = t^2 + 3sin(t).
3. Add the general solution of the homogeneous equation to the particular solution obtained in step 2, giving the complete solution.
Let's start with step 1:
1. Find the general solution to the associated homogeneous equation:
The homogeneous equation is 2y'' + 3y' + y = 0. To solve this, we assume a solution of the form y = e^(rt), where r is a constant to be determined. Substitute this into the equation:
2(e^(rt))'' + 3(e^(rt))' + e^(rt) = 0
Differentiate twice and substitute back into the equation:
2r^2e^(rt) + 3re^(rt) + e^(rt) = 0
Factor out e^(rt):
e^(rt)(2r^2 + 3r + 1) = 0
For a non-trivial solution, 2r^2 + 3r + 1 = 0. Factor this quadratic equation:
(2r + 1)(r + 1) = 0
This gives two possible values for r: r = -1/2 and r = -1.
Thus, the general solution to the homogeneous equation is:
y_h(t) = c1e^(-t/2) + c2e^(-t),
where c1 and c2 are arbitrary constants.
Now, let's move on to step 2:
2. Find a particular solution to the non-homogeneous equation:
To find a particular solution, we can use the method of undetermined coefficients. Since the right-hand side of the equation contains t^2 and sin(t), we assume a particular solution of the form y_p(t) = At^2 + Bsin(t), where A and B are constants to be determined.
Substitute this particular solution into the non-homogeneous equation:
2(2A) + 3(2Atcos(t) - Bsin(t)) + At^2 + Bsin(t) = t^2 + 3sin(t)
Simplify and equate coefficients of like terms:
(4A + A)t^2 + (6A - B + B)t + (4A - 3B) = t^2 + 3sin(t)
This gives the following equations:
4A + A = 1 (coefficient of t^2)
6A - B + B = 0 (coefficient of t)
4A - 3B = 3 (coefficient of sin(t))
Solving these equations, we find A = 1/4 and B = 1/2.
Therefore, the particular solution is:
y_p(t) = (1/4)t^2 + (1/2)sin(t)
Finally, let's move on to step 3:
3. Add the general solution of the homogeneous equation to the particular solution:
The complete solution is given by the sum of the general solution of the homogeneous equation (y_h) and the particular solution (y_p):
y(t) = y_h(t) + y_p(t)
= c1e^(-t/2) + c2e^(-t) + (1/4)t^2 + (1/2)sin(t),
where c1 and c2 are arbitrary constants. This is the complete solution to the given second-order differential equation.