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,) 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)=?

Step-by-step explanation:

Given is a differntial equation ,where x can take any positive value

One of the solution is

Let us assume the second solution

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

plug these in the DE

Put w=u'

xw'+11w=0

g(x) = x² + 2

step-by-step explanation:

it is given that the graphs are the same shape.

the only difference is that the red graph is 2 units higher than the blue graph.

that is to say that the red graph is simply the blue graph that has been moved vertically by 2 units in the positive y direction.

hence

g(x) = f(x) + 2

or

g(x) = x² + 2

i'm not sure i quite understand, could you provide more information?