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I am just begin to study integral equations, in which i come with following problem regarding second kind Volterra non-linear integro-differential equation, $$u'(x)=-1+\int_{0}^{x}u^{2}(t)dt$$ with the initial condition $\displaystyle u_{0}(x)=-x$. I want to know the exact solution of above problem i don't know how to proceed.... Thanks in advance....

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We derive $$u''(x)=u(x)^2$$

By trial and error I found that $u$ could be $$u(x)=\frac6{(x+k)^2}$$

where $k$ is a constant that can be easily computed from your integro-differential equation (I obtained $k=\sqrt 6$). I don't know if there are more solutions.

By the other hand, I don't understand that initial condition: $u_0(x)=x$. Isn't that a function definition? What is exactly $u_0$?.

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$u'(x)=-1+\int_0^xu^2(t)~dt$

$u''(x)=u^2(x)$

According to http://www.wolframalpha.com/input/?i=u%22%3Du%5E2,

$u(x)=\sqrt[3]6~\wp\left(\dfrac{x+C_1}{\sqrt[3]6};0,C_2\right)$

But I don't know how to substitute $u'(0)=-1$ .

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