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We know that y1(x)=x3 is a solution to the differential equation x2D2y+5xDy−21y=0 forx∈(0,[infinity]).

Use the method of reduction of order to find a second solution tox2D2y+5xDy−21y=0 for x∈(0,[infinity]).(a) After you reduce the second order equation by making the substitution w=u′, you get a first order equation of the form w′=f(x,w)=(b) y2(X)=?

Respuesta :

Answer:

Step-by-step explanation:

Given is a differntial equation [tex]x^2 y"+5xy'-21y=0[/tex],where x can take any positive value

One of the solution is

[tex]y_1 = x^3[/tex]

Let us assume the second solution [tex]y_2 = u x^3[/tex]

Differentiate this y2 two times and plug in the DE to reduce the order

[tex]y_2' = u'x^3 +3x^2 u\\y_2" = u"x^3+3x^2 u'+3x^2u'+6x u\\      = u"x^3+6x^2u'+6xu[/tex]

plug these in the DE

[tex]u"x^5+3x^4 u'+3x^4u'+6x^3 u+5u'x^4 +15x^3 u-21ux^3=0\\\\u"x^5+3x^4 u'+3x^4u'++5u'x^4 =0\\xu''+11u'=0[/tex]

Put w=u'

xw'+11w=0

[tex]y_2=ux^3[/tex]

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