dy/dt = -(t - 5) / (y + 2): can't find general solution

[khauna]

I am having trouble with this differential equation.

\(\displaystyle dy/dt=-(t-5)/(y+2)\)

when you solve using separation of variables, you get an expression like:

\(\displaystyle y[(y/2)+2y]=(-t^2/2)+5t+C\)

I am confused on how to find the general solution of this.

 

[Deleted member 4993]

Re: problem

I am having trouble with this differential equation.

\(\displaystyle dy/dt=-(t-5)/(y+2)\)

when you solve using separation of variables, you get an expression like:

\(\displaystyle (y^2/2)+4y=(-t^2/2)+2t+C\)

I am confused on how to find the general solution of this.

By general solution do you mean you want to write explicitly

y = f(t) ?

If yes then treat your solution as quadratic in 'y' and the roots of that equation will be your general solution.

 

[khauna]

Re: problem

yes i want y(t) explicitly.

thats the part im confused about, how do you do the quadratic part...

I get two solutions for y because i factor out the y and get

\(\displaystyle y[(y/2)+2]=(-t^2/2)+5t+C\)
so i have
\(\displaystyle y=(-t^2/2)+5t+C\) and \(\displaystyle y=(-t^2)+10t-4C\)
...so does that mean i do two quadratics?
If that is what i am supposed to do, then I will get a constant C under the square root
and I do not know how to deal with that.

 

[galactus]

Re: problem

On the left side, complete the square. Then you can easily solve for y.

 

[khauna]

Re: problem

I thought the \(\displaystyle y^2\) coefficient had to be 1 to complete the square, in this case we have 1/2 so i cant use that right?

 

[Deleted member 4993]

Re: problem

Do you know what are the solutions of the following equation

\(\displaystyle Ay^2 \, + \, By \, + \, C \, = \, 0\)

In your case

\(\displaystyle A \, = \, \frac{1}{2}\)

\(\displaystyle B \, = \, 2\)

\(\displaystyle C \, = \,\frac{t^2}{2} \, - 5t \, - \, C_1\)

 

[galactus]

Re: problem

Multiply through by 2 and get rid of the 1/2, then you have a leading 1.

 

[cans?n]

dy(y+2)=dt(-t+5) then take integral of this y^2/2+2y=C-t^2/2+5t and multiply by 2 and u will acquire the square of (y+2).(y^2+4y)=2C-t^2+10t . y^2+4y+4=2C+4-t^2+10t and then y is written as t function.y+2=in square root C-t^2+10t and y=C-t^2+10t in square root.