Answer to Question #327262 in Differential Equations for n hasan

Question

Find the differential equation whose solution is 𝑦 = 𝐴𝑥 + 𝐵𝑥².

Answer 1

Let's find $y'$ and $y''$ :

$y=Ax+Bx^2$

$y'=A+2Bx$ (1)

$y''=2B$ (2)

From (2) we have:

$B=\frac{y''}{2}$

Putting B into (1) we obtain:

$y'=A+2\frac{y''}{2}x=A+y''x$

$A=y'-y''x$

Substituting A and B into the origin equation we get:

$y=(y'-y''x)x+\frac{y''}{2}x^2$

$\frac{x^2}{2}y''-xy'+y=0$

Answer: $\frac{x^2}{2}y''-xy'+y=0$ .