Roisincleary
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Don’t know how to start any of Q5
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Roisincleary
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Am I completely wrong here?
D
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You are almost correct - the second constant should be C2
Roisincleary
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Oh of course hahaha
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You have correctly identified the characteristic roots, and so (as Subhotosh Khan pointed out), your homogeneous solution is thus:
[MATH]y_h(t)=c_1e^t+c_2e^{2t}[/MATH]
So, now what form must your particular solution take?
Roisincleary
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I’m afraid I’m not sure
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Because the RHS of the ODE is:
[MATH]g(t)=t^2[/MATH]
We may therefore assume the particular solution has the form:
[MATH]y_p(t)=At^2+Bt+C[/MATH]
Can you use the method of undetermined coefficients to find the parameters in the particular solution?
[MATH]g(t)=t^2[/MATH]
We may therefore assume the particular solution has the form:
[MATH]y_p(t)=At^2+Bt+C[/MATH]
Can you use the method of undetermined coefficients to find the parameters in the particular solution?
Roisincleary
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Let me check:
[MATH]y_p(t)=At^2+Bt+C[/MATH]
[MATH]y_p'(t)=2At+B[/MATH]
[MATH]y_p''(t)=2A[/MATH]
[MATH](2A)-3(2At+B)+2(At^2+Bt+C)=t^2[/MATH]
[MATH](2A)t^2+(2B-6A)t+(2A-3B+2C)=1t^2+0t+0[/MATH]
This implies:
[MATH]2A=1[/MATH]
[MATH]2B-6A=0[/MATH]
[MATH]2A-3B+2C=0[/MATH]
Solving this system, there results:
[MATH](A,B,C)=\left(\frac{1}{2},\frac{3}{2},\frac{7}{4}\right)\quad\checkmark[/MATH]
So, what is your particular solution, and then by the principle of superposition, what is the general solution to the given ODE?
[MATH]y_p(t)=At^2+Bt+C[/MATH]
[MATH]y_p'(t)=2At+B[/MATH]
[MATH]y_p''(t)=2A[/MATH]
[MATH](2A)-3(2At+B)+2(At^2+Bt+C)=t^2[/MATH]
[MATH](2A)t^2+(2B-6A)t+(2A-3B+2C)=1t^2+0t+0[/MATH]
This implies:
[MATH]2A=1[/MATH]
[MATH]2B-6A=0[/MATH]
[MATH]2A-3B+2C=0[/MATH]
Solving this system, there results:
[MATH](A,B,C)=\left(\frac{1}{2},\frac{3}{2},\frac{7}{4}\right)\quad\checkmark[/MATH]
So, what is your particular solution, and then by the principle of superposition, what is the general solution to the given ODE?
D
Deleted member 4993
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So your (homogeneous + particular) solution is:
y(t) = c1et + c2e2t + 1/2 * t2 + 3/2 * t + 7/4
Now apply boundary conditions to solve for c1 & c2
y(t) = c1et + c2e2t + 1/2 * t2 + 3/2 * t + 7/4
Now apply boundary conditions to solve for c1 & c2
HallsofIvy
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You were able to do the problems before #5? The only difference between 4 and 5 is that the equations in #5 are "non-homogeneous". They have a term, f(t), that depends on x only, no Y or derivatives of y.
For linear equations, we can get the general solution the entire equation by adding any one solution of the entire equation to the general solution to the "associated homogeneous equation" (dropping f(x)). So the problem becomes one of find a single solution to the entire equation.
There are two methods of doing that. The simpler is the "method of undetermined coefficients". IF f(t) is "of the kind we expect as solutions to linear des with constant coefficients (polynomials, sine or cosine, exponentials, and multiples of those) which is the case with all problems in #5. Experience with des allows you to guess what kind of function it should be, up to constants.
For example, in (a) \(\displaystyle f(t)= t^2\), a quadratic function, so we try \(\displaystyle y(t)= At^2+ Bt+ C\), the general quadratic. Then \(\displaystyle y'= 2At+ B\) and \(\displaystyle y''= 2At\).
Now put this into the equation, \(\displaystyle \frac{d^2y}{dt^2}- 3\frac{dy}{dt}+ 2y= 2At+ B- 3(2At+ B)+ 2(At^2- 3Bt+ C)= 2At^3+ (-4A- 6B)t- 2B+ 2C= t^2\). In order that two polynomials be the same for all t, the coefficients must be the same: 2A= 1, -4A- B= 0, -2B+ 2C= 0. A= 1/2, -2A- B= 0 so B= -1, and C= B= -1.
The general solution to the associated homogeneous equation, \(\displaystyle \frac{d^2y}{dt^2}- 3\frac{dy}{dt}+ 2y= 0\), which has characteristic equation [tex\r^2- 3r+ 2= (r- 2)(r+ 1)= 0[/tex], is \(\displaystyle y(t)= Ae^{2t}+ Be^{-t}\) so the general solution to the entire equation is \(\displaystyle y(t)= Ae^{2t}+ Be^{-t}+ (1/2)t^2- t- 1\).
For linear equations, we can get the general solution the entire equation by adding any one solution of the entire equation to the general solution to the "associated homogeneous equation" (dropping f(x)). So the problem becomes one of find a single solution to the entire equation.
There are two methods of doing that. The simpler is the "method of undetermined coefficients". IF f(t) is "of the kind we expect as solutions to linear des with constant coefficients (polynomials, sine or cosine, exponentials, and multiples of those) which is the case with all problems in #5. Experience with des allows you to guess what kind of function it should be, up to constants.
For example, in (a) \(\displaystyle f(t)= t^2\), a quadratic function, so we try \(\displaystyle y(t)= At^2+ Bt+ C\), the general quadratic. Then \(\displaystyle y'= 2At+ B\) and \(\displaystyle y''= 2At\).
Now put this into the equation, \(\displaystyle \frac{d^2y}{dt^2}- 3\frac{dy}{dt}+ 2y= 2At+ B- 3(2At+ B)+ 2(At^2- 3Bt+ C)= 2At^3+ (-4A- 6B)t- 2B+ 2C= t^2\). In order that two polynomials be the same for all t, the coefficients must be the same: 2A= 1, -4A- B= 0, -2B+ 2C= 0. A= 1/2, -2A- B= 0 so B= -1, and C= B= -1.
The general solution to the associated homogeneous equation, \(\displaystyle \frac{d^2y}{dt^2}- 3\frac{dy}{dt}+ 2y= 0\), which has characteristic equation [tex\r^2- 3r+ 2= (r- 2)(r+ 1)= 0[/tex], is \(\displaystyle y(t)= Ae^{2t}+ Be^{-t}\) so the general solution to the entire equation is \(\displaystyle y(t)= Ae^{2t}+ Be^{-t}+ (1/2)t^2- t- 1\).
Roisincleary
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Okay, the general solution to the ODE is:
[MATH]y(t)=c_1e^t+c_2e^{2t}+\frac{1}{2}t^2+\frac{3}{2}t+\frac{7}{4}[/MATH]
Now, to determine the parameters for the IVP, we use the given conditions:
[MATH]y(0)=c_1+c_2+\frac{7}{4}=1[/MATH]
[MATH]y(1)=c_1e+c_2e^2+\frac{1}{2}+\frac{3}{2}+\frac{7}{4}=2[/MATH]
Or:
[MATH]c_1+c_2=-\frac{3}{4}[/MATH]
[MATH]ec_1+e^2c_2=-\frac{7}{4}[/MATH]
Suppose we multiply the first equation by \(-e\) so that our system becomes:
[MATH]-ec_1-ec_2=\frac{3}{4}e[/MATH]
[MATH]ec_1+e^2c_2=-\frac{7}{4}[/MATH]
Adding the equations, we obtain:
[MATH](e^2-e)c_2=\frac{3e-7}{4}[/MATH]
[MATH]c_2=\frac{3e-7}{4e(e-1)}[/MATH]
Can you proceed?
[MATH]y(t)=c_1e^t+c_2e^{2t}+\frac{1}{2}t^2+\frac{3}{2}t+\frac{7}{4}[/MATH]
Now, to determine the parameters for the IVP, we use the given conditions:
[MATH]y(0)=c_1+c_2+\frac{7}{4}=1[/MATH]
[MATH]y(1)=c_1e+c_2e^2+\frac{1}{2}+\frac{3}{2}+\frac{7}{4}=2[/MATH]
Or:
[MATH]c_1+c_2=-\frac{3}{4}[/MATH]
[MATH]ec_1+e^2c_2=-\frac{7}{4}[/MATH]
Suppose we multiply the first equation by \(-e\) so that our system becomes:
[MATH]-ec_1-ec_2=\frac{3}{4}e[/MATH]
[MATH]ec_1+e^2c_2=-\frac{7}{4}[/MATH]
Adding the equations, we obtain:
[MATH](e^2-e)c_2=\frac{3e-7}{4}[/MATH]
[MATH]c_2=\frac{3e-7}{4e(e-1)}[/MATH]
Can you proceed?