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Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 4, Problem 17

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Problem

In each of problems 6-17, find a differential equation whose solution is the given n-parameter family.

\log(y) = c_1 x^2 + c_2

Solution

There are two constants so let’s take two derivatives.

First derivative.

\frac{y\prime}{y} = 2 c_1 x

y\prime = 2 c_1 x y

c_1 = \frac{y\prime}{2xy}

Second derivative.

y\prime\prime = 2 c_1 y + 2 c_1 x y\prime

Let’s substitute c_1 into the second derivative.

y\prime\prime = 2 \frac{y\prime}{2xy} y + 2 \frac{y\prime}{2xy} x y\prime

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

xyy\prime\prime = yy\prime + x (y\prime)^2

xyy\prime\prime - yy\prime - x (y\prime)^2 = 0

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 4, Problem 11

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Problem

In each of problems 6-17, find a differential equation whose solution is the given n-parameter family.

y = \sqrt{c_1 x^2 + c_2}

Solution

There are two constants, so it’s a good idea to take two derivatives and see what you have.

y\prime = \frac{c_1 x}{\sqrt{c_1 x^2 + c_2}}

y\prime\prime = \frac{c_1}{\sqrt{c_1 x^2 + c_2}} - \frac{c_1^2 x^2}{(c_1 x^2 + c_2)^{\frac{3}{2}}}

Notice that y is the denominator in both derivatives.

y\prime = \frac{c_1 x}{y}

y\prime\prime = \frac{c_1}{y} - \frac{c_1^2 x^2}{y^3}

Notice we got rid of c_2. Now we just need to get rid of c_1.

Let solve y\prime for c_1.

c_1 = \frac{yy\prime}{x}.

Substitute.

y\prime\prime = \frac{yy\prime}{xy} - \frac{(\frac{yy\prime}{x})^2 x^2}{y^3} = \frac{y\prime}{x} - \frac{(y\prime)^2}{y}

xyy\prime\prime + x (y\prime)^2 - y y\prime = 0

Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 4, Problem 9

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Problem

In each of problems 6-17, find a differential equation whose solution is the given n-parameter family.

r = \theta \tan(\theta + c)

Solution

I found it easier to simplify the input.

z = \theta + c

We’ll need at least one derivative.

r\prime = \theta \sec^2(z) + \tan(z)

Our goal here is to get rid of z, which would get rid of c.

Notice that because r = \theta \tan(\theta + c), we know that \tan(z) = \frac{r}{\theta}. Let’s substitute.

r\prime = \theta \sec^2(z) + \frac{r}{\theta}

Multiply everything by \theta.

\theta r\prime = \theta^2 \sec^2(z) + r

Remember that \sec^2(z) = \tan^2(z) + 1.

\theta r\prime = \theta^2( \tan^2(z) + 1 ) + r

\theta r\prime = \theta^2 \tan^2(z) + \theta^2 + r

Once again, \tan(z) = \frac{r}{\theta}.

\theta r\prime = \theta^2 ( \frac{r}{\theta} )^2 + \theta^2 + r

\theta r\prime = r^2 + \theta^2 + r

We have freed the equation from c. The differential equation is \theta r\prime = r^2 + \theta^2 + r.