This method requires one solution to be known. This is another method to solve a non-homogenous second order differential equation

Problem: Find the general solutionsolution for $x^2y''-7xy'+16y=0$​​ given that $x^4$​ is a solution

Solution: Use the method of reduction of order

We need to find the second solution $y_2$ where $y_2(t)=v(t)y_1(t)$

• $y_2 = ux^4$​​
• $y_2'= 4x^3u + u'x^4$​
• $y_2''=12x^2u + u'8x^3 + u''x^4$

Plug these derivatives in to the DE and simplify it

$x^6u'' + x^5u'=0$: A second order differential equation. Use $w=u'$​ and substitute $u$​ with $w$​ to get $x^6w' + x^5w=0$. Divide this by $x^5$ to simplify the equation to $xw'+w=0$.

We solved this differential equation in Problem 2, so $w=\frac{c_1}{x}$​​​​. Integrate this again to get $u$​​ since $w=u'$​​ and we need $u$​​, so $\int \frac{c_1}{x} dx=c_1ln(x)+c_2$​​. Therefore, $u=c_1ln(x) + c_2$​​ and so $y_2 = (c_1ln(x)+c_2)x^4$​​. We can get rid of the constants by setting $c_1=1$​​ and $c_2=0$​​​ and get $y2=ln(x)$​

General solution is $y = c_1y_1 + c_2y_2$​. Plug in to get the answer $y=c_1x^4 + c_2ln(x)x^4$​​