What is a solution to the differential equation #y'+y/x^2=2/x^3#?

2 Answers
Jul 17, 2016

#y = 2 (1/x + 1) +C e^{ 1/x} #

Explanation:

#y'+y/x^2=2/x^3#

let's try separating,...., #y'=2/x^3 - y/x^2# ie NOT separable, so in wrong category

BUT we can still do it! by writing it in this form we look for the Integrating Factor #eta(x)#

#y'+ color{blue}{1/x^2} y =2/x^3#

So here #eta(x) = e^{ int color{blue}{ 1/x^2 }\ dx} = e^{ - 1/x}#

#e^{ - 1/x} y'+ e^{ - 1/x} color{blue}{1/x^2} y =e^{ - 1/x}2/x^3#

or

#d/dx (y e^{ - 1/x} ) =e^{ - 1/x}2/x^3#

#int \ d/dx (y e^{ - 1/x} ) \ dx =int \ e^{ - 1/x}2/x^3 \ dx#

So

#y e^{ - 1/x} = color{red}{int \ e^{ - 1/x}2/x^3 dx} qquad triangle#

we'll try IBP on the red bit

#int \ e^{ - 1/x}2/x^3 dx#

#= int \ 1/x^2 e^{ - 1/x} * 2/x \ dx#

#= int \ d/dx( e^{ - 1/x}) * 2/x \ dx#

which by IBP: #int uv' = uv - int u' v#

#= e^{ - 1/x}2/x - int \ e^{ - 1/x} d/dx(2/x) \ dx#

#= e^{ - 1/x}2/x +2 int \ e^{ - 1/x} (1/x^2) \ dx#

#= e^{ - 1/x}2/x + 2 e^{ - 1/x} +C#

going back to #triangle#
#implies y e^{ - 1/x} = 2 e^{ - 1/x} (1/x + 1) +C#

#implies y = 2 (1/x + 1) +C e^{ 1/x} #

Jul 17, 2016

#y(x)=C_3e^{1/x}+2 (1 + 1/x)#

Explanation:

This is a linear homogeneus differential equation. Its solution can be assembled as

#y(x) = y_h(x) +y_p(x)#

where

#y'_h(x)+(y_h(x))/x^2=0# and
#y'_p(x)+(y_p(x))/x^2=2/x^3#

The homogeneus solution is straightforward. Grouping

#(y'_h(x))/(y_h(x))=-1/x^2->log_e y_h(x) = C_0+1/x->y_h(x)=C_1e^{1/x}#

For the obtention of #y_p(x)# we will use the Lagrange's variation of constants technique.

Supposing that #y_p(x) = C(x)e^{1/x}# and substituting in the complete equation we obtain

#C'(x) = (2e^{-1/x})/x^3->C(x)=2 e^(-1/x) (1 + 1/x) + C_2#

so

#y_p(x) = 2 (1 + 1/x) + C_2e^{1/x}#

Finally

#y(x) = (C_1+C_2)e^{1/x}+2 (1 + 1/x) = C_3e^{1/x}+2 (1 + 1/x)#