if dy/dx = y+3>0 & lny (0)=2,then y(ln2) is equal to :
a)7
b)5
c)13
d)-2
plzz solve & explain this question..!

We don't do homework. Please post YOUR answer and someone will come along and correct/explain it.

Something must be wrong with your question.

Is that:

If \frac{dy}{dx} = y+3>0 and ln_y (0)=2,then y(ln2) is equal to :

?

First, I never encountered this: \frac{dy}{dx} = y+3>0

And second, ln_y (0)=2 is false since the log of 0 to the base of any number gives 1.

Please, post the exact question, and your attempt at it.

The first part is a differential equation. Y(x) is a function. The equation says this:

\frac {dy(x)}{dx}=y(x)+3

which means that

y(x)=Ae^x-3,

where A is some unknown coefficient. Since the problem also states that the derivative is greater than 0, that means that the coefficient A must be a positive number.

The next part says

\ln {\( y(0) \)}=2,

but y(0) is simply A-3, so we can plug that into the equation:

\ln {\( A-3 \)}=2,

A-3=e^2

A=e^2+3

That means

y(x)=\( e^2+3\)e^x-3

So, to find y(ln(2)), we simply plug that in for x:

y \( \ln(2) \)=\( e^2+3\)e^{\ln(2)}-3=2\( e^2+3\)-3=2e^2+3

That doesn't match any of your multiple choice answers. I'm not sure if that means I made some silly mistake, or if it means that you copied the problem incorrectly, or if it means I'm misinterpreting what you wrote. :confused:

Unknown008, do you agree?

JC: I think you have interpreted the OP's question correctly. If the problem had stated that ln(y(0)) = ln(2) instead of ln(y(0)) =2, then one of the multiple choices would work out.

Good point EB! Now that you point it out, it seems clear that the OP must have just copied down the problem incorrectly. Hopefully they'll check back in at some point to verify.

Ah! Now I understand that better. I would never have guessed :p

I have done so little differential equations :(