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Consider the following second-order differential equation:
y``- 4y` +3y = 2t - 3t2,The particular solution of the differential equations is?
- a)-2-2t-t2
- b)-2t-t2
- c)2t-3t2
- d)-2-2t-3t2
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider the following second-order differential equation:y``- 4y` +3...
given y``- 4y` +3y = 2t - 3t2.
⇒ (D2 - 4D+3y = 2t - 3t2)
By the definition of particular solution-
verifying options, option (a) satisfies,
(D2-4D+3) (-2-2t - 3t2) = -2 + 8 + 8t - 6 - 6t 3t 2 = 2t - 3t2
∴(A) is constant
Most Upvoted Answer
Consider the following second-order differential equation:y``- 4y` +3...
To find the particular solution of the given second-order differential equation, we can use the method of undetermined coefficients. This method assumes that the particular solution has a similar form to the non-homogeneous term on the right-hand side of the equation.
First, let's write the given differential equation in standard form:
y'' - 4y' + 3y = 2t - 3t^2
Now, let's assume that the particular solution has the form:
y_p = At + Bt^2
Substituting this assumed form into the differential equation, we get:
y_p'' - 4y_p' + 3y_p = 2t - 3t^2
Differentiating y_p with respect to t:
y_p' = A + 2Bt
Differentiating y_p' with respect to t:
y_p'' = 2B
Now, substituting these derivatives back into the differential equation, we get:
2B - 4(A + 2Bt) + 3(At + Bt^2) = 2t - 3t^2
Simplifying the equation:
-4At + (2B + 3At) + 3Bt^2 + 2B = 2t - 3t^2
Collecting like terms:
(-4A + 2B + 3A)t + (3B)t^2 + 2B = 2t - 3t^2
Comparing coefficients, we have the following equations:
-4A + 2B + 3A = 2 (coefficient of t)
3B = -3 (coefficient of t^2)
2B = 0 (constant term)
Solving these equations, we find:
A = -2
B = 0
Therefore, the particular solution is:
y_p = -2t
Thus, the correct answer is option A: -2 - 2t - t^2.
First, let's write the given differential equation in standard form:
y'' - 4y' + 3y = 2t - 3t^2
Now, let's assume that the particular solution has the form:
y_p = At + Bt^2
Substituting this assumed form into the differential equation, we get:
y_p'' - 4y_p' + 3y_p = 2t - 3t^2
Differentiating y_p with respect to t:
y_p' = A + 2Bt
Differentiating y_p' with respect to t:
y_p'' = 2B
Now, substituting these derivatives back into the differential equation, we get:
2B - 4(A + 2Bt) + 3(At + Bt^2) = 2t - 3t^2
Simplifying the equation:
-4At + (2B + 3At) + 3Bt^2 + 2B = 2t - 3t^2
Collecting like terms:
(-4A + 2B + 3A)t + (3B)t^2 + 2B = 2t - 3t^2
Comparing coefficients, we have the following equations:
-4A + 2B + 3A = 2 (coefficient of t)
3B = -3 (coefficient of t^2)
2B = 0 (constant term)
Solving these equations, we find:
A = -2
B = 0
Therefore, the particular solution is:
y_p = -2t
Thus, the correct answer is option A: -2 - 2t - t^2.
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