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I want to ask simple questions.

$$ dy/dx + p(x)y=q(x)$$

In courses, I see always $$p(x)y$$ but what if $$ p(x)y^2$$ or $$p(x)y^3$$ or more ?

It doesn't change anything I guess but I want to ask anyway.

I mean what if

$$ dx/dx + y^3=5x$$ will p(x) = 1 again ?

Tryingtogetsome
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  • The ODE $dy/dx+p(x)y=q(x)$ is linear but $dy/dx+p(x)y^{n>1}=q(x)$ is not linear. – A. P. Apr 23 '22 at 23:04
  • yes, I don't get it. I am trying to solve one question and it says use the given formula (not this formula but same with these only there is $$e^p(x)dx... $$ . Why they ask this question if I can't solve with these formula? – Tryingtogetsome Apr 23 '22 at 23:16
  • My bad, I found the problem. looks like i had to change the equation – Tryingtogetsome Apr 23 '22 at 23:36

1 Answers1

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The integrating factor method is applicable only to linear differential equations of the form $$ {dy \over dx} + P(x) y = Q(x) \tag{1} $$

For solving (1), we define the integrating factor $$ \mu = \exp\left( \int P(x) dx \right) $$ and find the general solution of (1) with the formula $$ y \, \mu = \int \mu Q(x) \, dx + C $$ where $C$ is an integration constant.

However, if you consider the ODE as $$ {dy \over dx} + P(x) y^2 = Q(x) \tag{2} $$ or $$ {dy \over dx} + P(x) y^3 = Q(x) \tag{3} $$ then (2) and (3) are nonlinear differential equations due to the presence of $y^2$ or $y^3$ in them. In such cases, you cannot use the integration factor method to solve them.

Dr. Sundar
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