Question

Use the method of variation of parameters to determine the general solution of the given differential equation. NOTE: Use c1, c2, and c3 as arbitrary constants. y''' + y' = tan(t), -?/2 < t < ?/2. Suppose the general solution is y(t) = yc(t) + Y(t), where yc(t) = c1 + c2 cos(t) + c3 sin(t) is the homogeneous solution and Y(t) = -ln(cos(t)) + 1 - sin(t) ln(tan(t) + sec(t)) is the particular solution.

          Use the method of variation of parameters to determine the general solution of the given differential equation.

NOTE: Use c1, c2, and c3 as arbitrary constants.

y''' + y' = tan(t), -?/2 < t < ?/2.

Suppose the general solution is y(t) = yc(t) + Y(t), where

yc(t) = c1 + c2 cos(t) + c3 sin(t)

is the homogeneous solution and

Y(t) = -ln(cos(t)) + 1 - sin(t) ln(tan(t) + sec(t))

is the particular solution.

Use the method of variation of parameters to determine the general solution of the given differential equation.

NOTE: Use c1, c2, and c3 as arbitrary constants.

y”' + y' = tan(t), -?/2 < t < ?/2.

Suppose the general solution is y(t) = yc(t) + Y(t), where

yc(t) = c1 + c2 cos(t) + c3 sin(t)

is the homogeneous solution and

Y(t) = -ln(cos(t)) + 1 - sin(t) ln(tan(t) + sec(t))

is the particular solution.

Added by Steven B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

03/31/2023

Step 1

Therefore, the complementary solution is: $$y_c(t) = C_1 + C_2 e^{-t}$$ Now, we use the method of variation of parameters to find the particular solution. We assume the particular solution is in the form: $$Y(t) = u_1(t) + u_2(t)e^{-t}$$ Taking the first Show more…

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Use the method of variation of parameters to determine the general solution of the given differential equation. NOTE: Use c1, c2, and c3 as arbitrary constants. y''' + y' = tan(t), -pi/2 < t < pi/2 Suppose the general solution is y(t) = yc(t) + Y(t), where yc(t) = c1 + c2 cos(t) + c3 sin(t) is the homogeneous solution and Y(t) = -ln(cos(t)) + 1 - sin(t) ln(tan(t) + sec(t)) is the particular solution.