Answer to Question #284193 in Differential Equations for ASAP

Question

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

Answer 1

1.

$y_2=y_1v$

$y_2=t^2v,y_2'=2tv+t^2v',y_2''=2v+2tv'+2tv'+t^2v''=2v+4tv'+t^2v''$

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

$t^2v''=0$

$v=t$

$y_2=t^3$

2.

Wronskian:

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

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

then:

$W=cx$

$y_2'sinx^2-2y_2xcosx^2=cx$

$y_2=uv,y_2'=u'v+uv'$

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

$u'v=cx/(sinx^2)$

$v'-2vxcotx^2=0$

$dv/v=2xcotx^2dx$

$lnv=ln(sinx^2)$

$v=sinx^2$

$du=cxdx/(sinx^4)$

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

$y_2=\frac{4ln(sinx)+csc^2x(x(cscxcos3x-3cotx)-1)}{6}sin^2x$