Differential Equations

Question

Solve the given initial-value problem. Give the largest interval I over which the solution is defined.

(x+1)dy/dx + y = ln x,  y(1) = 10

y=?

Answer 1

(x+1)*y' + y = ln x

y' + (1/(x+1)) * y = (ln x)/(x+1) <--- divides both sides by x+1

The integrating factor is more informative than useful:

exp ( - integral( (x+1)^(-1)) =

exp ( - ln(x+1)) =

1/ exp( ln(x+1)) =

1/(x+1)

This returns the same ODE, so that means it can be re-written

in terms of the differential operation

Note that Dx [ ( x + 1) * y ] = (x+1) * dy/dx + y, by product rule, which in fact is the left side of the ODE

Dx [ (x+1) * y ] = ln x

Integrating:

(x+1)y = integral (ln x)

(x+1)y = x(ln x - 1) + C <--- can be integrated by parts (IBP) or table lookup

y = [x(ln x - 1) + C] / (x+1)

Per the initial condition:

10 = [ 1 * [ ln 1 - 1] + C]/(1+1)

10 = [ 1 * [ 0-1] + C]/2

20 = 1 * (-1) + C

20 = C - 1

C =21

The solution is

y = [x(ln x - 1) + 21] / (x+1)

CHECK:

Quotient Rule says:

dy/dx = {(x+1) [ x( 1/x) + (ln x - 1) ] - [x(ln x - 1) + 21] } / {x+1}^2

= { (x+1) *ln x - xln x + x - 21 ] } / {x+1}^2 <---- x cancels 1/x cancels -1; distribute x across (lnx-1)

and then all signs of second glob change

= { xlnx + lnx - xln x +x - 21}/{x+1}^2

= {lnx +x - 21}/(x+1)^2

Plugs this into the ODE:

dy/dx * (x+1) = {lnx +x - 21}/(x+1)

Adding this to y, the denominator, already common,

is x+1, while the numerator is:

[x(ln x - 1) + 21] + {lnx +x - 21}

the x cancels out.....

21 cancels out.......

xlnx + ln x remains in the numerator

lnx (x+1) remains in the numerator

x+1 cancels out

ln x remains which shows the solution is correct

the domain of the solution is x not equal to -1 and x>0.

the latter condition satisfies the former, so the domain is (0,infinity)