Answer to Question #233800 in Differential Equations for Phyroehan

Question

Find the general/particular solution of the following Differential Equations.

(Integrable Combinations)

1.) (xdy-ydx)/(x^2) = x^3 dx

Answer 1

$\frac{{xdy - ydx}}{{{x^2}}} = {x^3}dx \Rightarrow xdy - ydx = {x^5}dx \Rightarrow x\frac{{dy}}{{dx}} - y = {x^5} \Rightarrow \frac{{dy}}{{dx}} - \frac{y}{x} = {x^4} \Rightarrow y' - \frac{y}{x} = {x^4}$

Let

$y = u(x)v(x) \Rightarrow y' = u'v + uv'$

Then

$u'v + uv' - \frac{{uv}}{x} = {x^4} \Rightarrow u'v + u\left( {v' - \frac{v}{x}} \right) = {x^4}$

Let

$v' - \frac{v}{x} = 0 \Rightarrow \frac{{dv}}{{dx}} = \frac{v}{x} \Rightarrow \frac{{dv}}{v} = \frac{{dx}}{x} \Rightarrow \ln v = \ln x \Rightarrow v = x$

Then

$u'x = {x^4} \Rightarrow u' = {x^3} \Rightarrow u = \frac{{{x^4}}}{4} + C$

Then

$y = uv = \left( {\frac{{{x^4}}}{4} + C} \right)x = \frac{{{x^5}}}{4} + Cx$

Answer: $y = \frac{{{x^5}}}{4} + Cx$