Answer to Question #220816 in Calculus for Unknown346307

Question

Given that y = x is a particular solution of the differential

equation y''-x2y'

+ xy = 0. Find its general solution.

Answer 1

We make the substitution $y=y_1u=xu$

Then

$$y'=u+xu'$$

$$y''=u'+u'+xu''=2u'+xu''$$

Substitute

$$2u'+xu''-x^2u-x^3u'+x^2u=0$$

$$xu''+(2-x^3)u'=0$$

Let $p=u'.$ Then $u''=p'.$

$$xp'+(2-x^3)p=0$$

$$\dfrac{dp}{p}=\dfrac{x^3-2}{x}dx$$

Integrate both sides

$$\int\dfrac{dp}{p}=\int\dfrac{x^3-2}{x}dx$$

$$\ln|p|=\dfrac{x^3}{3}-2\ln|x|+\ln C$$

$$p=\dfrac{c_1e^{x^3/3}}{x^2}$$

$$u=\int \dfrac{c_1e^{x^3/3}}{x^2}dx$$

$$y=x\int \dfrac{c_1e^{x^3/3}}{x^2}dx$$