## solve `dy/dx=(-1xy)/(1+x^2` with y(0)=2

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### solve `dy/dx=(-1xy)/(1+x^2` with y(0)=2

The question says solve `dy/dx=(-1xy)/(1+x^2` with y(0)=2 as initial condition.
Explore and know. That is asked. barnamah

Posts: 102
Joined: Mon 14 March, 2011 08:37

### Re: solve `dy/dx=(-2xy)/(1+x^2)` with y(0)=2

we have the initial condition y(0)=2 and the differential equation equations `dy/dx=(-2xy)/(1+x^2)`
as you can see this equation is separable so we have separate all x with dx on one side all y with dy on the other side.
multiply both side by dx and we have
`(dx)*dy/dx=(-2xy)/(1+x^2)*(dx)`
`dy=(-2xy)/(1+x^2)*(dx)`

to get ys with dy, divide both side by y or `1/y`so we have
`1/y*dy=(-2xy)/(1+x^2)*(dx)*1/y` and we have:
`1/y*dy=(-2x)/(1+x^2)*(dx)` which is
`dy/y=(-2x)/(1+x^2)*(dx)` Now integrate
`int1/ydy=int(-2x)/(1+x^2)dx`
on the left side we have ln(y) and on the right we have:

`int(-2x)/(1+x^2)dx`

using u=`1+x^2` and du=2xdx we we have:
`-int1/udu` and after integration it is -ln(u)+c and substitute u we have:
`-ln(1+x^2)+c`
To simplify it we know -ln(A)=(-1)ln(A) and `ln(A)^-1` so `-ln(1+x^2)+c` becomes `ln(1+x^2)^(-1)=ln(1/(1+x^2))` so our equation becomes
`ln|y|=ln(1/(1+x^2))+c` now do the `e` and
`e^(ln|y|)=e^(ln(1/(1+x^2))+c)`
we have `|y|=e^(ln(1/(1+x^2))+c)`
`|y|=e^(ln(1/(1+x^2)))*e^c` and because `e^c` is a constant we can call is k. so we have:
`y=+-e^(ln(1/(1+x^2)))*e^c` by removing the absulut values, we put `+-` and then the `+-` go whit the constant, e cancels the ln and we have:
`y=1/(1+x^2)*k`

Now use the initial condition y(0)=2
`2=1/(1+0^2)*k`
`2=1/(1)*k`
`2=k`

so we found the k to be 2 and out equation `y=1/(1+x^2)*k`
`y=2/(1+x^2`

and that is the solution. Integr88

Posts: 71
Joined: Tue 15 March, 2011 13:27 