Question

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

Answer 1

`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`.

Answer 2

`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`