Applying Partial Fractions to Solve Laplace Step Function Problems

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step function - laplace...

Homework Statement


y"+y = f(t)

f(t) = 1, t<pi/2
0, pi/2<=t<infinity

The Attempt at a Solution



i now have L{y} = \frac{1-e^(-(pi/2)s)+s}{s(s^2+1)}

but how do i separate them and finish the problem?

thanks
 
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Use partial fractions like always.
\frac{1}{s(s^2+ 1)}= \frac{A}{s}+ \frac{Bs+ C}{s^2+ 1}

Find A, B, and C.
 


HallsofIvy said:
Use partial fractions like always.
\frac{1}{s(s^2+ 1)}= \frac{A}{s}+ \frac{Bs+ C}{s^2+ 1}

Find A, B, and C.

so no i have f(t) = 1-cos(t)+sin(t) - L{\frac{e^(pi*t/2)}{s(s^2+1)}}

how do i get the last laplace?
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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