Use the Laplace transform to solve the given initial-value problem. y + 4yet, y(0) - 2 Step 1 To use the Laplace transform o solve the given initial value problem, we first take the transform of each member of the differential equation +4y=e³t. The strategy is that the new equation can be solved for (y) algebraically. Once solved, transforming back to an equation for y gives the solution we need to the original differential equation. {{0}+ We now recall the following, where a and b are constants. + £{4y)=L(0²³²) By Theorem 7.2.2.: . Using linearity of L: z(ay) = ax{v} • By Theorem 7.1.1.: (000) == 0 Applying these gives the following result. sz(y)-y(0)+( -sz(y)-y(0) 1)x(x) = dt

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Use the Laplace transform to solve the given initial-value problem.
y' + 4y = e³t, y(0) = 2
Step 1
dy
To use the Laplace transform to solve the given initial value problem, we first take the transform of each member of the differential equation + 4y = e³t. The strategy is that the new equation can be solved for {y} algebraically. Once solved, transforming back to an equation
dt
for y gives the solution we need to the original differential equation.
+ L{4y} = L{e³t}
We now recall the following, where a and b are constants.
= s£{y} − y(0)
By Theorem 7.2.2.:
• Using linearity of L:
• By Theorem 7.1.1.:
L{ay} = a£{y}
{b} = 5=b
Applying these gives the following result.
] ) £{v} =
SL{y} - Y(0) +
Transcribed Image Text:Tutorial Exercise Use the Laplace transform to solve the given initial-value problem. y' + 4y = e³t, y(0) = 2 Step 1 dy To use the Laplace transform to solve the given initial value problem, we first take the transform of each member of the differential equation + 4y = e³t. The strategy is that the new equation can be solved for {y} algebraically. Once solved, transforming back to an equation dt for y gives the solution we need to the original differential equation. + L{4y} = L{e³t} We now recall the following, where a and b are constants. = s£{y} − y(0) By Theorem 7.2.2.: • Using linearity of L: • By Theorem 7.1.1.: L{ay} = a£{y} {b} = 5=b Applying these gives the following result. ] ) £{v} = SL{y} - Y(0) +
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