Solve dy/dx=(y+x^2-2)/(x+1) ,y(0)=2 ??

2 Answers
Mar 31, 2018

#dy/dx=0#

Explanation:

The key to solving this problem is understanding what it means. The expression #y(0)=2# tells us that y=2 when x=0. Knowing this, it's fairly simple to solve the equation by substituting #x# and #y# with their values.

#dy/dx=(y+x^2-2)/(x+1)#

#dy/dx=(2+0^2-2)/(0+1)=0/1=0#

This tells us that the derivative of y with respect to #x# equals #0# when #y = 2# and #x = 0#.

Mar 31, 2018

#y(x) = (x+1)^2-2(x+1)ln|x+1|+1#

Explanation:

#dy/dx = (y+x^2-2)/(x+1) implies#

# dy/dx - y/(x+1) = (x^2-2)/(x+1)#

This is a linear first order differential equation of the form

#dy/dx+P(x)y = Q(x)#

where #P(x) = -1/(x+1)# and #Q(x)=(x^2-2)/(x+1)#. It is well known that such an equation admits of an integrating factor given by

# exp(int P(x) dx) = exp(-int dx/(x+1)) = exp(-ln(x+1)) = 1/(x+1)#

Multiplying both sides of the equation by this factor gives us

#1/(x+1) dy/dx - y/(x+1)^2 = (x^2-2)/(x+1)^2#

Now, the left hand side is

# 1/(x+1) dy/dx + d/dx( 1/(x+1)) y = d/dx(y/(x+1)) #

On the other hand

# (x^2-2)/(x+1)^2 = (x^2color(red)(+2x+1)color(blue)(-2x-1)-2)/(x+1)^2#
#qquad = ((x+1)^2-2(x+1)-1)/(x+1)^2 = 1-2/(x+1)-1/(x+1)^2#

Thus, the equation becomes

# d/dx(y/(x+1)) = 1-2/(x+1)-1/(x+1)^2#

Integration with respect to #x# yields

# y/(x+1) = x-2ln|x+1|+1/(x+1)+C#

The constant of integration #C# can be determined by using the initial condition #y(0) = 2# :

#2/(color(red)0+1) = color(red)0-2ln|color(red)0+1|+1/(color(red)0+1)+C implies C=1#

Thus, the solution is

# y/(x+1) = x-2ln|x+1|+1/(x+1)+1#

or

#y(x) = (x+1)^2-2(x+1)ln|x+1|+1#