# Differential Equation: Initial Value Problem

Solve explicitly the initial value problem: $x^2y'+xy=2+x^2$, $y(1) = 2$

In the solution, they begin by dividing by $x^2$ to normalize the ODE, and multiply both RHS and LHS by the integrating factor x, yielding the integrable problem

$xy'+y=\frac{2}{x}+x, y(1)=2.$

Then they say by inspection, $(xy)'=\frac{2}{x}+x,$ ignoring the y on the LHS.

So, why do they ignore y? Doesn't y change with respect to x, meaning that the y on the LHS contributes to the equation?

## 1 Answer

Product rule:

$$(xy)' = x'y + xy' = y + xy'$$

using that

$$\frac{d}{dx} x = 1$$

• I don't think this answers the question "why do they ignore $y$". Though maybe it does because I can't understand what that's supposed to mean. $\huge\cdot$__$\huge\cdot$ – user137731 Nov 3 '16 at 3:16
• @Bye_World: you should think about it more. The OP's question states that he doesn't see why $(xy)'$ and $xy' + y$ are the same, thinking that the first one is "ignoring $y$". – Willie Wong Nov 3 '16 at 3:18
• Ah -- the "ignoring the $y$ on the LHS" is not part of what "they say". Got it. – user137731 Nov 3 '16 at 3:19
• Hum... punctuation in addition to italics would certainly have helped there. – Willie Wong Nov 3 '16 at 3:21
• Aha! thank you very much for insight. I've been staring at this equation for a good 5 minutes until I resorted to stack exchange. – Corp. and Ltd. Nov 3 '16 at 3:22