What is laplacetransform (1+x)e^{2x} ?

The answer to laplacetransform (1+x)e^{2x} is (s-1)/((s-2)^2)

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laplacetransform (1+x)e^{2x}

laplace transform

(1 + x) e^{2 x}

\frac{s - 1}{( s - 2 ) ^{2}}

Solution steps

L {e^{2 x} (x + 1)}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

(1 + x) e^{2 x} : f (x) = (1 + x), a = 2

= L {(1 + x)} (s - 2)

L {(1 + x)} : \frac{1}{s} + \frac{1}{s ^{2}}

= \frac{1}{( s - 2 )} + \frac{1}{( s - 2 ) ^{2}}

= \frac{1}{s - 2} + \frac{1}{( s - 2 ) ^{2}}

Simplify

\frac{1}{s - 2} + \frac{1}{( s - 2 ) ^{2}} : \frac{s - 1}{( s - 2 ) ^{2}}

= \frac{s - 1}{( s - 2 ) ^{2}}

\int_{0}^{5} 1 + 3 x d x

f (x) = 3 x, at x = 3

(4 x^{2} + 2 y^{2}) d x - x y d y = 0

\frac{d y}{d x} = x - 2 y + 12

f (x) = 5 x + 7

What is laplacetransform (1+x)e^{2x} ?
The answer to laplacetransform (1+x)e^{2x} is (s-1)/((s-2)^2)