What is inverse oflaplace (2s+3)/((s-7)^4) ?

The answer to inverse oflaplace (2s+3)/((s-7)^4) is e^{7t}t^2+(17e^{7t}t^3)/6

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inverse oflaplace (2s+3)/((s-7)^4)

inverse laplace

\frac{2 s + 3}{( s - 7 ) ^{4}}

e^{7 t} t^{2} + \frac{17 e ^{7 t} t ^{3}}{6}

Solution steps

L^{- 1} {\frac{2 s + 3}{( s - 7 ) ^{4}}}

Take the partial fraction of

\frac{2 s + 3}{( s - 7 ) ^{4}} : \frac{2}{( s - 7 ) ^{3}} + \frac{17}{( s - 7 ) ^{4}}

= L^{- 1} {\frac{2}{( s - 7 ) ^{3}} + \frac{17}{( s - 7 ) ^{4}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{2}{( s - 7 ) ^{3}}} + L^{- 1} {\frac{17}{( s - 7 ) ^{4}}}

L^{- 1} {\frac{2}{( s - 7 ) ^{3}}} : e^{7 t} t^{2}

L^{- 1} {\frac{17}{( s - 7 ) ^{4}}} : \frac{17 e ^{7 t} t ^{3}}{6}

= e^{7 t} t^{2} + \frac{17 e ^{7 t} t ^{3}}{6}

(3 y^{2} + 2 x) d x + (6 x y + 16 y^{3}) d y = 0

\int sin (x) - 5 cos (x) d x

y^{^{'}} - 4 y = 4 e^{6 t}

\int \frac{x}{1 - x ^{4}} d x

sin ((π x)^{7})

What is inverse oflaplace (2s+3)/((s-7)^4) ?
The answer to inverse oflaplace (2s+3)/((s-7)^4) is e^{7t}t^2+(17e^{7t}t^3)/6