# Thread: Simple Calc Question - don't know how to start

1. ## Simple Calc Question - don't know how to start

Find $y$ as a function of $x$ if

$\frac{dy}{dx} = y(2x+1)$

and $y = -3$ when $x = 0$.

Any idea on where to get started? I have no clue...

2. Originally Posted by eddie2042
Find $y$ as a function of $x$ if

$\frac{dy}{dx} = y(2x+1)$

and $y = -3$ when $x = 3$.

Any idea on where to get started? I have no clue...
This is a first order linear differential equation.

Since it is separable, integrate on both sides to find y, that is,

$\int \frac{dy}{y} = \int (2x+1) dx$

3. Originally Posted by harish21
This is a first order linear differential equation.

Since it is separable, integrate on both sides to find y, that is,

$\int \frac{dy}{y} = \int (2x+1) dx$
thanks though i must apologize. I made an edit on the first post. I said that $y = -3$ when $x = 3$. It should be when $x = 0$.

Thanks; i'll see if it works.

4. Originally Posted by eddie2042
thanks though i must apologize. I made an edit on the first post. I said that $y = -3$ when $x = 3$. It should be when $x = 0$.

Thanks; i'll see if it works.
$y=-3$ when $x=0$ will give you the value of the constant $C$ that is in your function. Integrate first to find an equation of y as a function of x. then plug in the given values of y and x, which will give you the value of the constant.

5. Originally Posted by harish21
$y=-3$ when $x=0$ will give you the value of the constant $C$ that is in your function. Integrate first to find an equation of y as a function of x. then plug in the given values of y and x, which will give you the value of the constant.
Ok here's what i did so far. Tell me if i went wrong somewhere

$\int\frac{dy}{d} = \int(2x +1)dx$

$ln |y| +C = x^2 + x + C$

... then what..? The C's cancel out.. if i'm not mistaken..

6. Originally Posted by eddie2042
Ok here's what i did so far. Tell me if i went wrong somewhere

$\int\frac{dy}{d} = \int(2x +1)dx$

$ln |y| +C = x^2 + x + C$

... then what..? The C's cancel out.. if i'm not mistaken..
The Cs are arbitrary constants, and when you do the integral of both the left/right side it is sufficient to have only one constant variable (i.e. only one C) either on the left side or right side of your equation.

To obtain y, put everything up to the e. This will bring down y due to the rule of logs and that is your answer (assuming you've done the rest correct)

7. Originally Posted by eddie2042
Ok here's what i did so far. Tell me if i went wrong somewhere

$\int\frac{dy}{d} = \int(2x +1)dx$

$ln |y| +C = x^2 + x + C$

... then what..? The C's cancel out.. if i'm not mistaken..
NO. Thats a bit too impulsive.

The value of constant is not same on both the sides. if you have written C on the left side, then express the constant on the right hand side with another letter. (say $C'$). After that, bring all the constants on the same side. this gives you

$ln |y| = x^2 + x + (C'-C)$

note that $C'-C$, the difference of two constants, is also a constant. For convenience, denote it with another letter.

then find y

8. Originally Posted by harish21
NO. Thats a bit too impulsive.

The value of constant is not same on both the sides. if you have written C on the left side, then express the constant on the right hand side with another letter. (say $C'$). The value of constant is NOT the same. After that, bring all the constants on the same side. this gives you

$ln |y| = x^2 + x + (C'-C)$

note that $C'-C$, the difference of two constants, is also a constant. For convenience, denote it with another letter.

then find y

ok great; thanks you two!

9. Originally Posted by eddie2042
ok great; thanks you two!
Hope you can finish up the problem now. Reply back if you have any problem(s).