I have difficulties to solve these two differential equations: 1) y &#x2032; </m

Grimanijd 2022-07-13 Answered
I have difficulties to solve these two differential equations:
1) y ( x ) = x y ( x ) x + y ( x ) with the initial condition y ( 1 ) = 1 .I'm arrived to prove that
y = x ( 2 e 2 ( ln x + c ) 1 )
but I don't know if it's correct. If it's right, how do I find the constant c? Because WolframAlpha says that the solution is y ( x ) = 2 x 2 + 1 x.
2) y ( x ) = 2 y ( x ) x 2 x y ( x ) . I'm arrived to prove that z 1 ( z + 1 ) 3 = e 2 c x 2 but I don't know if it's correct. If it's right, how do I explain z to substitute it in y = x z? Then, how do I find the constant c ?
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Answers (1)

zlepljalz2
Answered 2022-07-14 Author has 22 answers
y ( y + x ) = x y
y x + y = x y y
( x y ) = x 1 2 ( y 2 )
Integrate
x y = 1 2 x 2 1 2 y 2 + C
y 2 x 2 + 2 x y = C
Evaluate the constant:
y ( 1 ) = 1 C = 2
( y + x ) 2 = 2 ( x 2 + 1 )
Finally,
y ( x ) = 2 ( x 2 + 1 ) x
You are on the right track
y = x ( 2 e 2 ( ln x + c ) 1 )
y = x ( 2 e 2 c x 2 1 )
Note that e 2 c = k
y = 2 x 2 k x
y = 2 x 2 + 2 x
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Obtaining Differential Equations from Functions
dydx=x21
is a first order ODE,
d2ydx2+2(dydx)2+y=0
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for
y=ex(Acosx+Bsinx) and the steps that I followed are as follows.
dydx=ex(Acosx+Bsinx)+ex(Asinx+Bcosx)
=y+ex(Asinx+Bcosx) (1)
d2ydx2=dydx+ex(Asinx+Bcosx)+ex(AcosxBsinx)
=dydx+(dydxy)y
using the orginal function and (1). Finally,
d2ydx22dydx+2y=0,
which is the required differential equation.
Similarly, if the function is y=(Acos2t+Bsin2t), the differential equation that I get is
d2ydx2+4y=0
following similar steps as above.