Question

Please I need help with differential equation. Thank you

Solve: dt(y+y1/2)/(1-t)=dy(t+1)

1-Determine the interval of validity for the above solution.

2- Redo the above differential equation given that y(-1/2)=1

Answer 1

1. I suppose the ODE is supposed to be

`\mathrm dt(y+y^(1/2))/(1-t)=\mathrm dy(t+1)`

Solving for
`(\mathrm dy)/(\mathrm dt)` gives

`(\mathrm dy)/(\mathrm dt)=(y+y^(1/2))/(1-t^2)`

which is undefined when
`t=\pm1`. The interval of validity depends on what your initial value is. In this case, it's
`t=-\frac12`, so the largest interval on which a solution can exist is
`-1\le t\le1`.

2. Separating the variables gives

`(\mathrm dy)/(y+y^(1/2))=(\mathrm dt)/(1-t^2)`

Integrate both sides. On the left, we have

`\displaystyle\int(\mathrm dy)/(y^(1/2)(y^(1/2)+1))=2\int(\mathrm dz)/(z+1)`

where we substituted
`z=y^(1/2)` - or
`z^2=y` - and
`2z\,\mathrm dz=\mathrm dy` - or
`\mathrm dz=(\mathrm dy)/(2y^(1/2))`.

`\displaystyle\int(\mathrm dy)/(y^(1/2)(y^(1/2)+1))=2\ln|z+1|=2\ln(y^(1/2)+1)`

On the right, we have

`\frac1{1-t^2}=\frac12\left(\frac1{1-t}+\frac1{1+t}\right)`

`\displaystyle\int(\mathrm dt)/(1-t^2)=\frac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^(1/2)+C`

So

`2\ln(y^(1/2)+1)=\ln(1-t^2)^(1/2)+C`

`\ln(y^(1/2)+1)=\frac12\ln(1-t^2)^(1/2)+C`

`y^(1/2)+1=e^{\ln(1-t^2)^(1/4)+C}`

`y^(1/2)=C(1-t^2)^(1/4)-1`

I'll leave the solution in this form for now to make solving for
`C` easier. Given that
`y\left(-\frac12\right)=1`, we get

`1^(1/2)=C\left(1-\left(-\frac12\right)^2\right))^(1/4)-1`

`2=C\left(\frac54\right)^(1/4)`

`C=2\left(\frac45\right)^(1/4)`

and so our solution is

`\boxed{y(t)=\left(2\left(\frac45-\frac45t^2\right)^(1/4)-1\right)^2}`