# Math Help - First Order Differential Equations: Solving Particular and Geneal Solution

1. ## First Order Differential Equations: Solving Particular and Geneal Solution

How to do you solve these?

1) find an explicit general solution of the differential equation:
dy/dx= ysinx

2) Find an implicit general solution of the differential equation:
y^(3) dy/dx= (y^(4)+1) cosx

3) Find an explicit particular solution of the initial value problem:
x dy/dx - y= 2x^(2) y, y(1)= 1

4) Find an explicit particular solution of the initial value problem:
dy/dx= 6e^(2x-y), y(0)= 0

5) A pitcher of buttermilk initially at 25 Degree celcius is to be cooled by setting it on the front porch, where the temperature is 0 degree celcius. Suppose that the temperature of the buttermilk has dropped to 15 Degree Celcius after 20 minutes, when will it be 5 degree celcius?

2. Originally Posted by googoogaga
1) find an explicit general solution of the differential equation:
dy/dx= ysinx
This one's easy. ("Divide and conquer." )

$\frac{dy}{dx} = y sin(x)$

$\frac{dy}{y} = sin(x) dx$

$\int \frac{dy}{y} = \int sin(x) dx$

$ln(|y|) = -cos(x) + C$

$|y| = e^{-cos(x) + C}$

or perhaps you would prefer the form:
$|y| = e^{-cos(x)}e^{C} = Ae^{-cos(x)}$

In either form the exponential is always positive (note then that we need to require that the constant A is positive) so we may drop the absolute value bars:
$y = Ae^{-cos(x)}$

-Dan

3. Originally Posted by googoogaga
2) Find an implicit general solution of the differential equation:
y^(3) dy/dx= (y^(4)+1) cosx
This is the same idea as the first.

$y^3 \frac{dy}{dx} = (y^4 + 1)cos(x)$

$\frac{y^3}{y^4 + 1} dy = cos(x) dx$

$\int \frac{y^3}{y^4 + 1} dy = \int cos(x) dx$

For the LHS we may substitute $z = y^4 + 1$ ==> $dz = 4y^3 dy$, so

$\int \frac{y^3}{y^4 + 1} dy = \frac{1}{4} \int \frac{1}{z} = \int cos(x) dx$

$\frac{1}{4}ln(|z|) = sin(x) + C$

$|z| = Ae^{4sin(x)}$<-- Just for the record, $A = e^{4C}$

Again, we may drop the absolute value bars, but we must require that $z = y^4 + 1$ never be negative. But this is true anyway, so no worries!

$y^4 + 1 = Ae^{sin(x)}$

Since all the question asked for is an implicit relationship, you can be done with this line here.

-Dan

4. Originally Posted by googoogaga
3) Find an explicit particular solution of the initial value problem:
x dy/dx - y= 2x^(2) y, y(1)= 1
$x \frac{dy}{dx} - y = 2x^2$

I forget whose name is attached to this equation, but we may simplify it somewhat by the substitution
$x = e^t$

which implies
$dx = e^t dt$ ==> $\frac{d}{dx} = e^{-t}\frac{d}{dt}$

So the differential equation becomes:
$e^t \left ( e^{-t} \frac{dy}{dt} \right ) - y = 2e^{2t}$

$\frac{dy}{dt} - y = 2e^{2t}$

The characteristic equation for the homogeneous equation is
$m - 1 = 0$

Thus m = 1 ==> $y_h(t) = Ae^{t}$

The particular solution is provided by a term of the form $y_p(t) = Be^{ct}$. Substitution in the t version of the differential equation gives:
$Bce^{ct} - Be^{ct} = 2e^{2t}$

Thus c = 2 (by matching the exponents) and Bc - B = 2 (by matching the coefficients) thus B = 2 also.

So the solution to the t version of the differential equation is
$y(t) = Ae^{t} + 2e^{2t}$

Now we need to revert back to the x variable. $x = ln(t)$, so
$y(x) = Ae^{ln(x)} + 2e^{2ln(x)} = Ax + 2x^2$

Now we have the initial condition: $y(1) = 1$.
$y(1) = A(1) + 2(1)^2 = 2 + A = 1$

Thus $A = -1$

and the solution to the differential equation will be:
$y(x) = 2x^2 - x$

-Dan

5. Originally Posted by googoogaga
4) Find an explicit particular solution of the initial value problem:
dy/dx= 6e^(2x-y), y(0)= 0
This is done in the same manner as 1) and 2). I get that
$y(x) = ln(3e^{2x} - 2)$

Originally Posted by googoogaga
5) A pitcher of buttermilk initially at 25 Degree celcius is to be cooled by setting it on the front porch, where the temperature is 0 degree celcius. Suppose that the temperature of the buttermilk has dropped to 15 Degree Celcius after 20 minutes, when will it be 5 degree celcius?
Here we need to assume that the temperature change is a linear function (according to Newton's Law of Cooling it should be), so we have
$\frac{dT}{dt} = \text{constant} = \frac{25 ^oC - 15 ^oC}{-20~min} = -0.5 ^oC/min$

So the temperature starts as $25 ^oC$ and cools at a constant $0.5 ^oC/min$. How long will it take to drop to $5 ^oC$?

-Dan

6. ## First Order Differential Equations: Solving Particual and general Solution...

# 4 still isn't crystal clear. How do you get the time? I am not really sure what to do here. Could you further explain please? Thanks.

7. Originally Posted by googoogaga
4) Find an explicit particular solution of the initial value problem:
dy/dx= 6e^(2x-y), y(0)= 0
$\frac{dy}{dx} = 6e^{2x - y} = 6e^{2x}e^{-y}$

$e^y dy = 6e^{2x} dx$

$\int e^y dy = \int 6e^{2x} dx$

Can you take it from here?

-Dan

8. Originally Posted by googoogaga
How do you get the time? I am not really sure what to do here. Could you further explain please? Thanks.
I presume you mean for #5. There's not much else I can say without just giving you the answer (and this one's too easy for that.)

Let me put it this way:
You are looking at a plane of graph paper where time, t, is on the x-axis and temperature, T, is on the y-axis. We have two points on the function we wish to graph:
$(0~min, 25 ^oC)$
and
$(20~min, 15 ^oC)$

I told you the function connecting these points is linear, that is to say, a line. I found the slope of this line, $-0.5 ^oC/min$. Now you need to supply the following coordinate: What is the t value when the T value is $5 ^oC$? ie. Fill in the following t coordinate: $(t, 5 ^oC)$.

If you were asked this about the xy plane and a line connecting the points I am sure you would have no problem doing this.

I should note that in writing this I realized that $\frac{dT}{dt}$ is negative. I have fixed this in my original post on this question. Sorry for the error!

-Dan