Find the general solution?

Question

Find the general solution of y⁽⁴⁾-y=3t+cos t.

Answer: y=c₁e^t+c₂e^-t+c₃cos t+c₄sin t-3t-(1/4)t*sin t

r=1, -1, i, -i,

But how do I figure out the form of the particular solution?

Answer 1

Find the homogeneous solution first by writing the characteristic equation:

r^4 - 1 = 0

(r^2 + 1) ( r^2-1) = 0

r = 1, -1, i, -i

The homogeneous solution is of the form:

y_h = c₁e^t + c₂ e^-t + c₃ cos(t) + c₄ sin (t)

Particular solution can be solved in 2 steps since there are 2 forcing functions on the right hand side:

Step1 :

y⁽⁴⁾ - y = 3t

we can guess the particular solution to be of the form  Y = At+B

Y^' = A

Y^'' = 0

Y^''' = 0

Y⁽⁴⁾ = 0

 

Substituting in the expression above,

-(At + B)  = 3t

which means A= -3 and B = 0

Y  = -3t

Now to step 2 to find the particular solution for,

Step2:

 

y⁽⁴⁾- y = cos(t)

We can guess our  linearly independent solution should be of the form:

 

Y= At cos(t) + Bt sin(t)

 

 

 

Y' = Acos(t) + B sin(t) − At sin(t) + Bt cos(t)

Y'' = −Asin(t) + B cos(t) − Asin(t) + B cos(t) − At cos(t) − Bt sin(t)

=−2Asin(t) + 2B cos(t) − Y

Y''' = −2Acos(t) − 2B sin(t) − Y'

Y⁽⁴⁾ = 2Asin(t) − 2B cos(t) − Y''

= 4Asin(t) − 4B cos(t) + Y

Y⁽⁴⁾ − Y = cos(t)

4Asin(t) − 4B cos(t) + Y − Y = cos(t)

4Asin(t) − 4B cos(t) = cos(t)

Then,

A = 0 and B = −1/4, which means Y = −(1/4) t sin(t)

The particular solution is the sum of the particular solution for step1 and step2

Y_{P =} -3t-(1/4) t sin(t)

Thus our general solution now becomes:
y=c1et+c2e-t+c3cos t+c4sin t-3t-(1/4)t*sin t

 

 

Answer 2

Plugging y = e^rt into the differential equation y⁽⁴⁾-y = 0,

r⁴ - 1 = (r²+1)(r+1)(r-1) = 0

r = ±i, 1, -1

So, the homegeneous solution is y_h = c₁e^t+c₂e^-t+c₃cos t+c₄sin t

Since cos t is part of homegeneous solution,  the particular solution can be written as

y* = D tsin t - 3t

y*' = D(sin t + t cos t) - 3

y*'' = D(cos t + cos t - t sin t)

y*''' = D(-3sin t - t cos t)

y*'''' = D(-4cos t + tsin t)

So, y*'''' - y* = 3t + cos t becomes

D(-4cos t + tsin t) - D tsin t + 3t = 3t + cos t

-4Dcos t = cos t => D = -(1/4)

Answer: y = y_h + y* = c₁e^t+c₂e^-t+c₃cos t+c₄sin t-3t-(1/4)t*sin t

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Attn: Since sin t and cos t are in homegeneous solution, and y = t cos t leads no solution, y* = D tsin t - 3t  is assumed.