# Solution of an integral equation $\phi(x)+\int^1_0 xt(x+t)\phi(t)\,dt=x$ , $0 \le x \le 1$

Solve the following integral equation:

$\phi(x)+\displaystyle \int^1_0 xt(x+t)\phi(t)\,dt=x$ , $0 \le x \le 1$

I need to solve the integral equation above. Can anyone help me please?

• What method do you want to use? May 16, 2013 at 15:26

Hint: $$\phi(x)+x^2 \displaystyle \int_0^1t\phi(t)dt+x\int_0^1t^2\phi(t)=x$$ i.e. $$\tag{1}\phi(x)=c_1x^2+c_2x$$

Where $c_1 = - \displaystyle \int_0^1t\phi(t)dt \ \$ and $\ \ c_2=1- \displaystyle \int_0^1t^2\phi(t)dt$

Put $\phi(t)$ from $(1)$ in these integrals and calculate $c_1 \ , \ c_2$ from two equations you get

(two-equations two variables).

• You can use the \tag{} command to tag equations. Apparently, it doesn't accept \spadesuit as an input, sorry about that.
– Pedro
May 17, 2013 at 1:37

Related problems: (I). Here is a detailed solution that maybe someone benefits from it. Rearranging the equation as $$\phi(x)= x- \displaystyle \int^1_0 xt(x+t)\phi(t)\,dt \longrightarrow (1).$$

$$\phi(x) = x - x^2\int_{0}^{1}t\phi(t)\,dt - x \int_{0}^{1}t^2\phi(t) \,dt$$

$$\implies \phi(x) = x - x^2 c_1 - x c_2 \longrightarrow (2).$$

Now, just subs back in the integral equation $(1)$ and compare the coefficients of $x's$, you will get a system of 2 equations in $c_1$ and $c_2$

$$\frac{5}{4}c_{1}+\frac{1}{3}c_{2}= \frac{1}{3}$$

$$\frac{1}{5}c_{1}+\frac{5}{4}c_{2}= \frac{1}{4}.$$

Solving the above system gives

$$\left\{ c_{{1}}={\frac {80}{359}},c_{{2}}={\frac {59}{359}} \right\}.$$

Subs back in $(2)$ yields the solution

$$\phi \left( x \right) ={\frac {300}{359}}\,x-{\frac {80}{359}}\,{x}^{2 },$$

which can be checked by plugging back in $(1)$.

• @moderators: This down vote is a misleading. May 16, 2013 at 16:22
• @Downvoters: What's the down vote for? This is the second one. May 17, 2013 at 1:45
• This solution is correct, +1. I don't know why it was downvoted. But does @moderators really notify the moderators? And do they really have to be notified in such a case? What could they do anyway? May 17, 2013 at 1:45
• @julien: It is really a misleading down vote. I worked out the problem in details so it will be like a reference for people. May 17, 2013 at 1:48
• And what is the outcome of doing so? Has a moderator ever done anything about it? Are they notified by @moderators? What can they do anyway? I think you should ask yourself these questions first. May 17, 2013 at 1:56