Question

The homogeneous differential equation t^2y'' - 4ty' + 6y = 0 (t > 0) has two solutions given by y1(t) = t^2 and y2(t) = t^3. Using the method of Variation of Parameters, find the general solution of the nonhomogeneous equation t^2y'' - 4ty' + 6y = t^3 y = C1t^2 + C2t^3 - t^2 ln(t) + t^4 y = C1t^2 + C2t^3 + t^2 ln(t) + t^4 None of these answers y = C1t^2 + C2t^3 + t^3 ln(t) y = C1t^2 + C2t^3 + 1/6t^5

          The homogeneous differential equation
t^2y'' - 4ty' + 6y = 0 (t > 0) has two solutions given by
y1(t) = t^2 and y2(t) = t^3.
Using the method of Variation of Parameters, find the general solution of the
nonhomogeneous equation
t^2y'' - 4ty' + 6y = t^3
y = C1t^2 + C2t^3 - t^2 ln(t) + t^4
y = C1t^2 + C2t^3 + t^2 ln(t) + t^4
None of these answers
y = C1t^2 + C2t^3 + t^3 ln(t)
y = C1t^2 + C2t^3 + 1/6t^5

The homogeneous differential equation
t^2y” - 4ty' + 6y = 0 (t > 0) has two solutions given by
y1(t) = t^2 and y2(t) = t^3.
Using the method of Variation of Parameters, find the general solution of the
nonhomogeneous equation
t^2y” - 4ty' + 6y = t^3
y = C1t^2 + C2t^3 - t^2 ln(t) + t^4
y = C1t^2 + C2t^3 + t^2 ln(t) + t^4
None of these answers
y = C1t^2 + C2t^3 + t^3 ln(t)
y = C1t^2 + C2t^3 + 1/6t^5

Added by Jennifer C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ivan Kochetkov

Step 1

First, we have the homogeneous equation: t^2y'' - 4ty' + 6y = 0 The given solutions are: Y_1(t) = t^2 Y_2(t) = t^3 Now, we want to find the general solution of the nonhomogeneous equation: ty'' - 4ty' + 6y = 3 Using the method of Variation of Parameters, we Show more…

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The homogeneous differential equation t^2y" - 4ty' + 6y = 0 (t > 0) has two solutions given by y1(t) = t^2 and y2(t) = t^3. Using the method of Variation of Parameters, find the general solution of the nonhomogeneous equation t^2y" - 4ty' + 6y = t^3 y = C1t^2 + C2t^3 - t^2 ln(t) + t^4