Problems on differential equation

The key thing to know is that this is an example of a Cauchy-Euler equation .

You can visit the link and work out the details yourself if you like. If not, then:

Assume a solution of the form $y=x^m$ .

Differentiate it twice and substitute the expressions for $y'$ and $y''$ to get: $(m(m-1)+m+1)x^m=0$ .

As the relation holds for all $x$ , we must have: $(m(m-1)+m+1)=0$ i.e $m^2+1=0$ Thus $m=\pm i$ .

Using the fact that this differential equation is linear, we must have the general solution to be of the form: $y=Ax^i+Bx^{-i}$ for some arbitrary constants A and B which we determine using the given conditions.

We get $A+B=1$ from $f(1)=1$ and $A-B=0$ from $f'(1)=0$ .

Thus, $y=\frac{x^i+x^{-i}}{2}$

As for the final answer, note that $i=e^{i\frac{\pi}{2}}$ and thus, $i^i=e^{-\frac{\pi}{2}}$

Then, $f(i^i)=\frac{e^{i\frac{\pi}{2}}+e^{-i\frac{\pi}{2}}}{2}=0$ .

Note:

For anyone with qualms about differentiating $x^i$ , it is to be noted that $\frac{\mathrm{d} }{\mathrm{d} x}x^m=mx^{m-1}$ for complex m as well.

This is because the derivative of $x^m$ is basically computed by using the binomial expansion which holds for a complex index as well.

i just solved and found y=Bcos(lnx)