Question #284193

I.In each of Problems 23 through 30, use the method of reduction of order to find a second solution of the given differential equation.

1.  t^2y″ − 4ty′ + 6y = 0, y1(t) = t^2

2.  xy″ − y′ + 4x^3y = 0, x > 0; y_1(x) = sin x^2


1
Expert's answer
2022-01-05T13:41:14-0500

1.

y2=y1vy_2=y_1v

y2=t2v,y2=2tv+t2v,y2=2v+2tv+2tv+t2v=2v+4tv+t2vy_2=t^2v,y_2'=2tv+t^2v',y_2''=2v+2tv'+2tv'+t^2v''=2v+4tv'+t^2v''


t2(2v+4tv+t2v)4t(2tv+t2v)+6t2v=0t^2(2v+4tv'+t^2v'') − 4t(2tv+t^2v')+ 6t^2v = 0

t2v=0t^2v''=0

v=tv=t

y2=t3y_2=t^3


2.

Wronskian:

W=y1y2y1y2=ceP(x)dxW=y_1y_2'-y_1'y_2=ce^{-\int P(x)dx}

where P(x)=1/xP(x)=-1/x


then:

W=cxW=cx

y2sinx22y2xcosx2=cxy_2'sinx^2-2y_2xcosx^2=cx

y2=uv,y2=uv+uvy_2=uv,y_2'=u'v+uv'

(uv+uv)sinx22uvxcosx2=cx(u'v+uv')sinx^2-2uvxcosx^2=cx


uv=cx/(sinx2)u'v=cx/(sinx^2)

v2vxcotx2=0v'-2vxcotx^2=0


dv/v=2xcotx2dxdv/v=2xcotx^2dx

lnv=ln(sinx2)lnv=ln(sinx^2)

v=sinx2v=sinx^2


du=cxdx/(sinx4)du=cxdx/(sinx^4)

u=4ln(sinx)+csc2x(x(cscxcos3x3cotx)1)6u=\frac{4ln(sinx)+csc^2x(x(cscxcos3x-3cotx)-1)}{6}


y2=4ln(sinx)+csc2x(x(cscxcos3x3cotx)1)6sin2xy_2=\frac{4ln(sinx)+csc^2x(x(cscxcos3x-3cotx)-1)}{6}sin^2x


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