Find $\phi(\log 2)$ for the integral equation $\phi(x)=1-2x-4x^2+\int_0^x[3+6(x-t)-4(x-t)^2]\phi(t)dt$ .

Question

If $\phi$ is the solution of the integral equation $$\phi(x)=1-2x-4x^2+\int_0^x[3+6(x-t)-4(x-t)^2]\phi(t)dt$$

Then the value of $\phi(\log 2)$ is

(a). 2

(b). 4

(c). 6

(d). 8

I tried this and I get the solution is $23$ but I am not sure about it can anyone please solve this.Thank you

Answer 1

Differentiating three times gives that $$\phi’’’(x)=3\phi’’(x)+6\phi’(x)-8\phi(x).$$ Solve the characteristic equation and we find it has three distinct roots, so everything will be easy.

The result I get is that $\phi(x)=e^x$ so $\phi(\log(2))=2$.

Thanks to @lan for pointing out my stupid errors.

Answer 2

Just for completeness, you can also do this with the Laplace transform, which uses the property that the Laplace transform of a "causal convolution" $(f*g)(x):=\int_0^x f(x-t) g(t) dt$ is $F(s) G(s)$. Consequently taking the Laplace transform of both sides gives

$$\Phi(s)=F(s) + G(s) \Phi(s)$$

where $F(s)$ is the Laplace transform of $1-2x-4x^2$, i.e. $\frac{1}{s}-\frac{2}{s^2}-\frac{8}{s^3}$, and $G(s)$ is the Laplace transform of $3+6x-4x^2$, i.e. $\frac{3}{s}+\frac{6}{s^2}-\frac{8}{s^3}$. Thus

$$\Phi(s)=\frac{F(s)}{1-G(s)}=\frac{s^2-2s-8}{s^3-3s^2-6s+8}.$$

From here you can carry out a partial fraction decomposition and take an inverse Laplace transform. Actually, you don't even have to carry out a partial fraction decomposition in this particular problem, you can just factor the numerator and denominator and observe the cancellation.