Laplace transform solve integral equation

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Homework Statement



Screen_Shot_2013_02_20_at_8_20_19_PM.png


Homework Equations


The Attempt at a Solution



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The answer is y(t) = t^{4}+\frac{t^{6}}{30}

Don't know what to do next any advices please
 
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solve algebraically for L{y}

L{y}=L{t^4}+L{y}L{sin(t)}
 
lurflurf said:
solve algebraically for L{y}

L{y}=L{t^4}+L{y}L{sin(t)}

u meant put L{y} to the same side and then reverse transform? I'm try that way but still didnt get that answer
 
lurflurf said:
solve algebraically for L{y}

L{y}=L{t^4}+L{y}L{sin(t)}

Thanks lurflurf I got it now :)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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