Solving Leplace Transform of y"+4y=7t with y(0)=-1, y'(0)=3

  • Thread starter Thread starter hils0005
  • Start date Start date
  • Tags Tags
    Transform
hils0005
Messages
61
Reaction score
0

Homework Statement



y"+4y=7t , y(0)=-1, y'(0)=3


The Attempt at a Solution



L{y} + 4L{y} = 7L{t}

(s^2L{y} - sy(0) - y'(0)) + 4L{y} = 7L{t}

s^2L{y} - s(-1) - 3 + 4L{y} = 7(0!/s^(0+1))

L{y}(s^2 + 4) = 7(0!/s^(0+1)) - s + 3
= ((-s+3)s + 7)/s
=(-s^2 + 3s +7)/s

L{y}=(-s^2 + 3s + 7)/(s(s^2+4))

here is where I run into trouble, setting up the partial fractions

L{y}=-s^2 + 3s +7 = A/s + B/(s^2+4)

is that correct so far?
 
Physics news on Phys.org
You should check the laplace transform for 7t.

L{t^n} = (n!)/(s^(n+1))
therefore L{t} = L{t^1} = (1!/(s^(1+1))

L{7t} should look more like 7*(1/s^2) or (7/s^2)
 
but its t^1, so n=0, t^0+1=t^1 right?
 
Its t^1, so n = 1 when you plug into the transform for t^n.
 
Or if you want the Laplace transform of t^n in terms of gamma function:

L\{t^n\}=\frac{\Gamma(n+1)}{s^{n+1}}

\Gamma(n+1)=n\Gamma(n)=...=n!,n \in Z^+
\Gamma(1)=1
 
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...
Back
Top