Simple IVP with Laplace

Calculus Level 3

Solve the initial value problem y ′ ′ ( t ) + y ( t ) = t y''(t) + y(t) = t , { y ( 0 ) = 1 y ′ ( 0 ) = 1 \begin{cases} y(0) = 1\\y'(0) = 1\end{cases} by using the Laplace transform method.

1 − sin ⁡ ( t ) 1 - \sin(t) t + sin ⁡ ( t ) t + \sin(t) t + cos ⁡ ( t ) t + \cos(t) sin ⁡ 2 ( t ) \sin^2(t)

Ingrid Spangler by Ingrid Spangler , 23, Brazil

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1 solution

Ingrid Spangler
May 15, 2018

L{y''(t)} = S²Y(S) - SY'(0) - Y(0)

L{y(t)} = Y(S)

L{t} = 1 S ² \frac{1}{S²}

With the values Y(0) = 1 and Y'(0) = 1...

S²Y(S) - S - 1 + Y(S) = 1 S ² \frac{1}{S²}

(S² + 1)Y(S) = 1 S ² \frac{1}{S²} + S + 1

Y(S) = 1 S ² ( S ² + 1 ) \frac{1}{S²(S² + 1)} + S ( S ² + 1 ) \frac{S}{(S² + 1)} + 1 ( S ² + 1 ) \frac{1}{(S² + 1)}

Now solving partial fractions for the denominator of just the first one A S ² \frac{A}{S²} + B ( S ² + 1 ) \frac{B}{(S² + 1)}

A = -B

1 S ² \frac{1}{S²} - 1 ( S ² + 1 ) \frac{1}{(S² + 1)}

gives us three defined laplace transforms

L⁻¹{ 1 S ² ( S ² + 1 ) \frac{1}{S²(S² + 1)} } = t - sin(t)

L⁻¹{ S ( S ² + 1 ) \frac{S}{(S² + 1)} } = cos(t)

L⁻¹{ 1 ( S ² + 1 ) \frac{1}{(S² + 1)} } = sin(t)

answer: Y(S) = t + cos(t)

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