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Have you tried differentiating it?How do I solve the Volterra integral equation?

\[

f(x) = \sqrt{x} + \lambda\int_0^x\sqrt{xy}f(y)dy

\]

.

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The derivative isHave you tried differentiating it?

.

\[

f'(x) = \frac{1}{2\sqrt{x}} + \lambda\left(xf(x) + \int_0^1\frac{y}{2\sqrt{xy}}f(y)dy\right)

\]

How does this help?

Now you can replace the integral in the expression for the derivative by the expression for it you get from the original equation leaving you with a first order ordinary differential equation.The derivative is

\[

f'(x) = \frac{1}{2\sqrt{x}} + \lambda\left(xf(x) + \int_0^1\frac{y}{2\sqrt{xy}}f(y)dy\right)

\]

How does this help?

(and you should still have \(x\) as the upper limit in the integral, you will also find things easier if you take the \(x\) outside the integral)

.

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SoNow you can replace the integral in the expression for the derivative by the expression for it you get from the original equation leaving you with a first order ordinary differential equation.

(and you should still have \(x\) as the upper limit in the integral, you will also find things easier if you take the \(x\) outside the integral)

.

\[

\int_0^x\sqrt{y}f(y)dy = \frac{f(x) - \sqrt{x}}{\lambda\sqrt{x}}

\]

Plugging this in we get

\[

f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\lambda\sqrt{x}} + f(x)\left(x\lambda + \frac{1}{2\lambda\sqrt{x}}\right).

\]

IWhat am I suppose to do with the \(f'(x)\) equation? The solution to this ODE isnt trivial.

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Well for \(x>0\), I get:So

\[

\int_0^x\sqrt{y}f(y)dy = \frac{f(x) - \sqrt{x}}{\lambda\sqrt{x}}

\]

Plugging this in we get

\[

f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2\lambda\sqrt{x}} + f(x)\left(x\lambda + \frac{1}{2\lambda\sqrt{x}}\right).

\]

IWhat am I suppose to do with the \(f'(x)\) equation? The solution to this ODE isnt trivial.

\[f'(x)=f(x)\left[\frac{1}{2x}+\lambda x\right]\]

which assuming no (further) mistakes in the manipulations is of variables seperable type with initial condition \(f(0)=0\)

Note the derivative is:

\[\frac{d}{d\,x}\,f\left( x\right) = \frac{\lambda\,\int_{0}^{x}\sqrt{y}\,f\left( y\right) dy}{2\,\sqrt{x}}+\lambda\,x\,f\left( x\right) +\frac{1}{2\,\sqrt{x}}\]

.

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How did you get that expression?Well for \(x>0\), I get:

\[f'(x)=f(x)\left[-\frac{1}{2x}+\lambda x\right]\]

which assuming no mistakes in the manipulations is of variables seperable type with initial condition \(f(0)=0\)

.

By substituting \(\frac{f(x) - \sqrt{x}}{\sqrt{x}}\) for \(\lambda \int_0^x\sqrt{y}f(y)dy \) in the expression for the derivative.How did you get that expression?

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Shouldn't it be + not minus? Also, for +, the ODE evaluates toBy substituting \(\frac{f(x) - \sqrt{x}}{\sqrt{x}}\) for \(\lambda \int_0^x\sqrt{y}f(y)dy \) in the expression for the derivative.

\[

f(x) = C \sqrt{x} e^{\frac{\lambda x^2}{2}}.

\]

With the IC of \(f(0) = 0\), \(f(x) = C\cdot 0 = 0\).

Which suggests that we go back to the original integral equation and rewrite it in terms of \(h(x)=f(x)/\sqrt{x}\), which if all has gone well is:Shouldn't it be + not minus? Also, for +, the ODE evaluates to

\[

f(x) = C \sqrt{x} e^{\frac{\lambda x^2}{2}}.

\]

With the IC of \(f(0) = 0\), \(f(x) = C\cdot 0 = 0\).

\[h(x)=1+\lambda \int_0^x y\; h(y)\ dy\]

which gives initial condition \(h(0)=1\) and:

\[h'(x) = x h(x)\]

which I think results in \(C=1\).

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