First order differential equation

[arshad]

Can anyone help me solve this equation;

y + 3dy/dt = x
y=?

Please remeber it's dy/dt not dy/dx

Thanks!

[factor-fiction]

It's 40 years since I did this stuff but can't you integrate in terms of dt
So:
yt+c +3y=xt+d where c and d are constants
so
yt+3y= xt+d-c= xt+w (where w is a constant)
so
y(t+3)= xt+w
so y= (xt+w)/(t+3)

[factor-fiction]

On reflection I'm sure my answer is wrong . Have we got enough info?
Do we know whether x is a constant or a function of y or t? And is y a function of just t (f(t))or of t and x (f(x,t))?

[arshad]

Thanks factor30 for trying, but your solution is not correct.
't' is a variable & 'x' could be a constant or a function.

[derrynoose]

I suspect that we have been asked the impossible here.
It's doubtful whether the question could be done even if it were dy/dx much less dy/dt.

[factor-fiction]

Hello Arshad,

Derrynoose and I are sure this can't be solved without more info on whether y is also a function of x and whether x is a function of t.
Are you sure the question is printed correctly on AB?
What was the actual question posed- maybe there's an error in your working before this point?

In what context has this equation arisen?

We want to help but need more info.

[arshad]

Ok factor30


y + 3dy/dt = x

x=1 (given)

y=?
Please remeber it's dy/dt not dy/dx

Answer is y=1-exp^-t/3
but how ..? or this answer is correct?

[factor-fiction]

So are you saying that the equation is basically:
y+3dy/dt=1
If so, then the answer is correct.
Starting with teh answer and working back:

if y= 1 - exp^(-t/3)
then dy/dt=(t/3)exp^(-t/3)
[since the general rule is d(e^kx)/dx=ke^kx]

Substituting these values into [y+3dy/dt-1] we get
1-e^(-t/3)+te^(-t/3)=1, which is true, showing solution works

To solve it without knowing the answer you could try the general solution y= a+ be^(kx). This will lead you to the answer

[derrynoose]

If y + 3dy/dt = 1
3dy/dt = 1-y
dy/1-y = dt/3
Integrate
-ln(1-y) = t/3 + C
-t/3 =ln(1-y) + C Let C = lnk where k is a constant.
-t/3 = ln(1-y) + lnk
-t/3 = lnk(1-y)
exp(-t/3) = k(1-y)
exp(-t/3)/k = 1-y
y = 1 -exp(-t/3)/k = general solution, if my keyboarding hasn't let me down!

[arshad]

Thanks derrynoose, it makes sense

[Robb Phoenix]

y=1-e^-t/3/k

[Robb Phoenix]

damn!!!!