Question #327262

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


1
Expert's answer
2022-04-12T10:05:45-0400

Let's find yy' and yy'' :

y=Ax+Bx2y=Ax+Bx^2

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

y=2By''=2B (2)

From (2) we have:

B=y2B=\frac{y''}{2}

Putting B into (1) we obtain:

y=A+2y2x=A+yxy'=A+2\frac{y''}{2}x=A+y''x

A=yyxA=y'-y''x

Substituting A and B into the origin equation we get:

y=(yyx)x+y2x2y=(y'-y''x)x+\frac{y''}{2}x^2

x22yxy+y=0\frac{x^2}{2}y''-xy'+y=0

Answer: x22yxy+y=0\frac{x^2}{2}y''-xy'+y=0 .


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