How do I solve the following initial value problem x dy + (y - y^2 ln x) dx = 0, y(1) = 1/4?

Answer 1

Given equation : xdy + (y – y^2 ln x) dx = 0: y(1)=1/4 Solution y = 1/(lnx +1+3x)

Steps:

1.    Divide by dx:                                     xdy/dx + y – y^2 lnx = 0     

2.    Rewrite dy/dx = y’:                            xy’ + y – y^2 lnx = 0

3.    Rewrite in form of Bernoulli ODE:   y’ + y/x = (lnx/x) y^2

4.    Bernoulli genral ODE:                      y’ + P(x)y = Q(x) y^n solve, transform into linear form

5.    Substituion v = y^1-n:                       v’/(1-n) + P(x) v = Q(x)

6.    For given equation:                          P(x) = 1/x , Q(x) = lnx/x and n=2. Therefore v = y^(1-2)

7.    Transform equation (TE):                -v’ + v/x = lnx/x  Linear differential  equation of first order  

8.    Integrating factor :                            I(x) = e ^ ⌠P(x) dx Contact me for intermediate steps.

9.    Solution of transformed equation:    v = lnx + 1 + C₁x 

10.   v = y^-1:                                          y = 1/( lnx + 1 +C₁x), ¼ = 1/(ln1 +1 +C₁)

11. Given y(1) = ¼, substitute find C₁:   y = 1/( lnx + 1 +3x)

Cheers! Happy July 4th

Answer 2

This question is so old, my help may be no help at all...especially since I can't solve it completely.

You have x dy + y dx = y² ln x

(x dy + y dx)/(x²y²) = (ln x)/x² dx

D[-1/(xy)] = (lnx)/x² dx ... and hopefully the right side will integrate by parts.

I will work on this some more and follow if I get any further.

5/27/2019

In fact x^-2ln x does integrate by parts to -x^-1ln - (1/2)ln²x so that gives you the general solution and you can use the initial condition to evaluate the constant of integration.

Answer 3

Rewrite the equation as y' +y/x -y²/x •lnx =0 divide through by y² to get y'/y² +1/xy -lnx/x=0 Let n= 1/y so that dn/dx = -1/y²y' and rewrite the equation as - dn/dx +n/x -lnx/x =0 this becomes dn/dx -n/x = -lnx/x use an

integrating factor m= e^ln(1/x) or m= 1/x multiply each term in the equation by 1/x to give 1/xdn/dx -n/x² = -lnx/x²

so now d/dx(n/x) -lnx/x² n/x = - ∫ lnx/x² + C. Use integration by parts by letting u=lnx and dv= dx/x² in so doing the integral becomes lnx/x + 1/x + C and substituting n=1/y gives 1/yx =lnx/x +1/x +C or 1/y= lnx +1 +Cx

and y = 1/(lnx +Cx +1 using the initial condition y(1)=1/4 gives 1/4 = 1/( 0+ 1 +C) or C=3. So the solution is

y= 1/(lnx + 3x+1)