1. ## differential equation

I am quite confused with the following question. Would appreciate any help to get me started. Thanks
Find one set of values of the numbers a and b, such that the function y = e^ax + bln(x) is a solution of the differential equations given below.

y” – ay’ = 0
b’ + y – b/x = 0

2. Originally Posted by mellowdano
I am quite confused with the following question. Would appreciate any help to get me started. Thanks
Find one set of values of the numbers a and b, such that the function y = e^ax + bln(x) is a solution of the differential equations given below.

y” – ay’ = 0
b’ + y – b/x = 0
You are told that $\displaystyle y = e^{ax} + b\ln{x}$.

So $\displaystyle y' = ae^{ax} + \frac{b}{x}$

$\displaystyle y'' = a^2e^{ax} - \frac{b}{x^2}$.

So equation 1:

$\displaystyle y'' - ay' = 0$

$\displaystyle a^2e^{ax} - \frac{b}{x^2} - a\left(ae^{ax} + \frac{b}{x}\right) = 0$

$\displaystyle a^2e^{ax} - \frac{b}{x^2} - a^2e^{ax} + \frac{ab}{x} = 0$

$\displaystyle \frac{abx - b}{x} = 0$

$\displaystyle abx - b = 0$

$\displaystyle b(ax - 1) = 0$

This means either $\displaystyle b = 0$ or $\displaystyle a = \frac{1}{x}$.

Equation 2:

$\displaystyle b' + y - \frac{b}{x} = 0$...

Are you sure you copied this down correctly?

3. Thanks! And yes the second one has a mistake it should be
y' + y - b/x = 0

So i'm thinking for equation 2;
ae^(ax) + b/x + e^(ax) + b ln(x) – b/x = 0
ae^(ax) + e^(ax) + b ln(x) = 0
a^2 ln(x) + a ln(x) + b ln(x) = 0
a^2 + a + b = 0
a(a +1 + b) = 0
a = 0