# Math Help - differential equations

1. ## differential equations

Form the differential equation which has a general solution of:
y = Cx^2 + C^2
I understand what to do, take the first derivative, find a formula for C and plug that back into y, but I dont see how to take this derivative because of the two Constants. Do i need to take 2 derivatives? Thanks.

2. Originally Posted by davecs77
Form the differential equation which has a general solution of:
y = Cx^2 + C^2
I understand what to do, take the first derivative, find a formula for C and plug that back into y, but I dont see how to take this derivative because of the two Constants. Do i need to take 2 derivatives? Thanks.
Let $y(x) = Ax^2 + B$
This has two "arbitrary" constants. I'll get to that in a minute.

Then
$y^{\prime}(x) = 2Ax$
and
$y^{\prime \prime}(x) = 2A$

Now, if we try to construct something like
$ay^{\prime \prime} + by^{\prime} + y = f(x)$
we'll get exponential terms in the solution. But if we use:
$ax^2y^{\prime \prime} + bxy^{\prime} + cy = f(x)$
we'll get polynomials.

Since we have no polynomial terms in the solution with specific (that is to say, constrained) coefficients, $f(x) = 0$.

So input the solution $y(x) = Ax^2 + B$:
$ax^2(2A) + bx(2Ax) + c(Ax^2 + B) = 0$

So
$2aA + 2bA + cA = 0$
and
$cB = 0$

Thus $c = 0$ and $2aA + 2bA = 0$ which further gives $b = -a$, so the differential equation is
$ax^2y^{\prime \prime} - axy^{\prime} = 0$

For simplicity we may drop the "a":
$x^2y^{\prime \prime} - xy^{\prime} = 0$

Now, how to get the form of $y(x) = Ax^2 + B$ to $y(x) = Cx^2 + C^2$? Well, this is a second order differential equation, so we need two constraint conditions. We are free to choose what conditions to specify, so long as the system is not overconstrained. But note that we have one undetermined variable, so we need only one condition. I would recommend something like:
$y(0) = \frac{1}{4}(y^{\prime}(1))^2$

It's a bizarre constraint, but it works.

So one possible differential equation would be:
$x^2y^{\prime \prime} - xy^{\prime} = 0$
such that
$y(0) = \frac{1}{4}(y^{\prime}(1))^2$

-Dan

3. Originally Posted by topsquark
Let $y(x) = Ax^2 + B$
This has two "arbitrary" constants. I'll get to that in a minute.

Then
$y^{\prime}(x) = 2Ax$
and
$y^{\prime \prime}(x) = 2A$

Now, if we try to construct something like
$ay^{\prime \prime} + by^{\prime} + y = f(x)$
we'll get exponential terms in the solution. But if we use:
$ax^2y^{\prime \prime} + bxy^{\prime} + cy = f(x)$
we'll get polynomials.

Since we have no polynomial terms in the solution with specific (that is to say, constrained) coefficients, $f(x) = 0$.

So input the solution $y(x) = Ax^2 + B$:
$ax^2(2A) + bx(2Ax) + c(Ax^2 + B) = 0$

So
$2aA + 2bA + cA = 0$
and
$cB = 0$

Thus $c = 0$ and $2aA + 2bA = 0$ which further gives $b = -a$, so the differential equation is
$ax^2y^{\prime \prime} - axy^{\prime} = 0$

For simplicity we may drop the "a":
$x^2y^{\prime \prime} - xy^{\prime} = 0$

Now, how to get the form of $y(x) = Ax^2 + B$ to $y(x) = Cx^2 + C^2$? Well, this is a second order differential equation, so we need two constraint conditions. We are free to choose what conditions to specify, so long as the system is not overconstrained. But note that we have one undetermined variable, so we need only one condition. I would recommend something like:
$y(0) = \frac{1}{4}(y^{\prime}(1))^2$

It's a bizarre constraint, but it works.

So one possible differential equation would be:
$x^2y^{\prime \prime} - xy^{\prime} = 0$
such that
$y(0) = \frac{1}{4}(y^{\prime}(1))^2$

-Dan
This is confusing to me..Here is an example I can show you of the way I understand it:
y = cx^2 + 1
yprime = c2x ...therefore x = yprime/2x
we can rewrite y as y = (yprime/2x)(x^2 + 1)
y = 1/2yprime(x+1)
2y = yprime(x+2)
(2y + 2)/(x) = yprime
Is there a way you can do the previous problem like this? By finding what the constant is (or constants) then plugging it back into y? Thanks again.

4. Originally Posted by davecs77
This is confusing to me..Here is an example I can show you of the way I understand it:
y = cx^2 + 1
yprime = c2x ...therefore x = yprime/2x
we can rewrite y as y = (yprime/2x)(x^2 + 1)
y = 1/2yprime(x+1)
2y = yprime(x+2)
(2y + 2)/(x) = yprime
Is there a way you can do the previous problem like this? By finding what the constant is (or constants) then plugging it back into y? Thanks again.
Clever! And simpler than mine, no less.

However I'm going to make two changes to the above. First your solution is $y(x) = Cx^2 + C^2$ and I am going to use the derivative equation to eliminate C, not x.
$y = Cx^2 + C^2$

$y^{\prime} = 2Cx \implies C = \frac{y^{\prime}}{2x}$

Thus
$y = \left ( \frac{y^{\prime}}{2x} \right ) x^2 + \left ( \frac{y^{\prime}}{2x} \right ) ^2$

$y = \frac{1}{2}x y^{\prime} + \frac{1}{4x^2}(y^{\prime}) ^2$

I would then multiply both sides by $x^2$ since the solution doesn't have the restriction on x = 0.

$4x^2y = 2x^3 y^{\prime} + (y^{\prime}) ^2$
(I threw in an extra factor of 4 to clear the fraction and clean it up a little.)

-Dan

5. Originally Posted by topsquark
Clever! And simpler than mine, no less.

However I'm going to make two changes to the above. First your solution is $y(x) = Cx^2 + C^2$ and I am going to use the derivative equation to eliminate C, not x.
$y = Cx^2 + C^2$

$y^{\prime} = 2Cx \implies C = \frac{y^{\prime}}{2x}$

Thus
$y = \left ( \frac{y^{\prime}}{2x} \right ) x^2 + \left ( \frac{y^{\prime}}{2x} \right ) ^2$

$y = \frac{1}{2}x y^{\prime} + \frac{1}{4x^2}(y^{\prime}) ^2$

I would then multiply both sides by $x^2$ since the solution doesn't have the restriction on x = 0.

$4x^2y = 2x^3 y^{\prime} + (y^{\prime}) ^2$
(I threw in an extra factor of 4 to clear the fraction and clean it up a little.)

-Dan
Yeah that makes a lot more sense to me..but what if you have C1 and C2? (2 different constants) Can you find one of the C's and say both are equal to that or do you need to find what each C is equal to?

I have one problem showing that..
y = C1e^x + C2e^(2x)
I know to find yprime
yprime = C1e^x + C2e^(2x)

6. Originally Posted by davecs77
Yeah that makes a lot more sense to me..but what if you have C1 and C2? (2 different constants) Can you find one of the C's and say both are equal to that or do you need to find what each C is equal to?

I have one problem showing that..
y = C1e^x + C2e^(2x)
I know to find yprime
yprime = C1e^x + C2e^(2x)
If you have two constants [tex]C_1[tex] and $C_2$ you cannot simply say that they are equal. (I usually call them different variable names like A and B if I can. It avoids confusion. )

Also, note that
$y^{\prime} = C_1e^x + 2C_2e^{2x}$.

To get a differential equation for this, recall that I noted
$ay^{\prime \prime} + by^{\prime} + cy = f(x)$
will give exponential solutions. In this case, that's the form you are looking for.

So you have two exponentials with coefficients in the exponentials of 1 and 2. Thus these are the roots of the characteristic equation:
$(m - 1)(m - 2) = 1 \cdot m^2 - 3 \cdot m + 2 = 0$

So your differential equation will be of the form:
$1 \cdot y^{\prime \prime} + (-3) \cdot y^{\prime} + 2 \cdot y = f(x)$
(The coefficients of the derivatives are the same as the coefficients of the terms in the characteristic equation.)

Again, since we have no terms in the solution with specific coefficients, $f(x) = 0$.

$y^{\prime \prime} - 3y^{\prime} + 2y = 0$