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$\frac{dy}{dx}+P(x)y=0$

  1. Show that if f and g are solutions to the differential equation where $c_1$ and $c_2$ are arbitrary constants then $c_1f$ & $c_2g$ are also solution to the differential equation

My attempt Since this equation is separable therefore,

$P(x)dx+\frac{dy}{y}=0$

By This differential equation must be exact!

$\frac{\partial P(x)}{\partial y}=\frac{\partial y^{-1}}{\partial x}=0$

$\int_{} P(x)=f+g+c_1$

$\int_{} \frac{1} {y} =lny+c_2$

I know this is lousy. Can someone guide me?

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  • $\begingroup$ Why don't you try to prove that $\frac{d}{dx} \left( c_1 f(x) + c_2g(x) \right) + P(x) (c_1f(x) +c_2g(x)) = 0$ holds? $\endgroup$ – Demophilus Jun 7 '17 at 15:58
  • $\begingroup$ It can be that way? $\endgroup$ – Crazy Jun 7 '17 at 15:59
  • $\begingroup$ I thought proof must start from the given question! The above is the exact question from a textbook. $\endgroup$ – Crazy Jun 7 '17 at 16:00
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    $\begingroup$ Of course, that is what is asked: If $\frac{d}{dx} f(x) + P(x) f(x) = 0$ and $\frac{d}{dx} g(x) + P(x) g(x) = 0$, show that the above holds. This shouldn't be hard since the derivative is a linear operator. $\endgroup$ – Demophilus Jun 7 '17 at 16:01
  • $\begingroup$ Thanks man! Now I see that. $\endgroup$ – Crazy Jun 7 '17 at 16:04
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It's simpler than all that. If $y(x) = f(x)$ satisfies the differential equation $\dfrac{dy}{dx} + P(x)y = 0$, then we know this equation is true: $$ \dfrac{df}{dx} + P(x) f = 0 \tag{*}$$

Using this given information, your goal is to show that $y(x) = c_1 f(x)$ satisfies the differential equation $\dfrac{dy}{dx} + P(x)y = 0$. In other words, you want to show this equation is true: $$ \dfrac{d}{dx}(c_1 f(x)) + P(x) c_1 f = 0$$

Well, it just follows from equation (*) above if we multiply both sides by $c_1$.

Similar reasoning for $c_2 g(x)$ (although it's weird to me that you're asked to essentially do the same exercise twice, since it really is literally the same thing using the names $c_2$ and $g$ instead of $c_1$ and $f$).

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  • $\begingroup$ It's a bit mysterious to me as well why both $f$ and $g$ are mentioned. The OP and one of the comments seems to assume the two functions should be added, but I don't see a justification for adding the two. $\endgroup$ – Χpẘ Jun 7 '17 at 22:05
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By hypothesis,

$$f'(x)+P(x)f(x)=0$$ and $$g'(x)+P(x)g(x)=0.$$

Then by linearity of the derivative operator,

$$c_1f'(x)+c_2g'(x)+c_1P(x)f(x)+c_2P(x)g(x)=0,$$ $$(c_1f(x)+c_2g(x))'+P(x)(c_1f(x)+c_2g(x))=0.$$

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