-aux.2.2.02 y=\phi(x)=(1-x^2)^{-1} is a solution IVP

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In summary: Think of the ODE as:y^{-2}\,dy=2x\,dxNow integrate, using the power rule on both sides...what do you get?Note: When you separated variables and divided through by $y^2$, you eliminated the trivial solution $y\equiv0$. As the initial value is not on that solution, you can ignore it, but we should be aware of any solution we eliminate during the process of solving.Think of the ODE as:y^{-2}\,dy=2x\,dxNow integrate, using the power rule on both sides...what do you get?Note: When you separated variables and divided
  • #1
karush
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$\tiny{2.3.2}$
1000
$\textsf{Given: }$
$$\displaystyle y'=2xy^2, \quad y(0)=1$$
$\textit{Show that $y=\phi(x)=(1-x^2)^{-1}$ is a solution of the initial value problem}$
\begin{align*}\displaystyle
y'&=2xy^2\\
\frac{dy}{y^2}&=2x \\
y&=\color{red}{\frac{1}{(c_1-x^2)}}
\end{align*}
ok I went into confusion after 2x??
red is W|F
 
Last edited:
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  • #2
Think of the ODE as:

\(\displaystyle y^{-2}\,dy=2x\,dx\)

Now integrate, using the power rule on both sides...what do you get?

Note: When you separated variables and divided through by $y^2$, you eliminated the trivial solution $y\equiv0$. As the initial value is not on that solution, you can ignore it, but we should be aware of any solution we eliminate during the process of solving.
 
  • #3
MarkFL said:
Think of the ODE as:

\(\displaystyle y^{-2}\,dy=2x\,dx\)

Now integrate, using the power rule on both sides...what do you get?

Note: When you separated variables and divided through by $y^2$, you eliminated the trivial solution $y\equiv0$. As the initial value is not on that solution, you can ignore it, but we should be aware of any solution we eliminate during the process of solving.

$\displaystyle \int \frac{1}{y^2} \, dy =\int 2x \, dx$
$\displaystyle -\frac{1}{y} = x^2 +c$

ok not sure if this is really the power rule but
also why is not c given on both as I noticed on examples

so

$\displaystyle-\frac{1}{x^2 +c}=y=\phi(x)$

why the introduction of $\phi(x)$ ??
 
  • #4
The power rule is:

\(\displaystyle \int u^r\,du=\frac{u^{r+1}}{r+1}+C\)

It appears you've properly applied it. Are you asking why there is not a constant of integration on both sides after you integrated?
 
  • #5
MarkFL said:
The power rule is:

\(\displaystyle \int u^r\,du=\frac{u^{r+1}}{r+1}+C\)

It appears you've properly applied it. Are you asking why there is not a constant of integration on both sides after you integrated?

Kinda
Some examples seem to do a
Magic disappearing act on the c's
 
  • #6
karush said:
Kinda
Some examples seem to do a
Magic disappearing act on the c's

There's really only the need to put a constant on 1 side...consider:

\(\displaystyle f(y)\,dy=g(x)\,dx\)

Now, let's integrate, and put a constant on both sides:

\(\displaystyle F(y)+c_1=G(x)+c_2\)

Subtract $c_1$ from both sides:

\(\displaystyle F(y)=G(x)+c_2-c_1\)

Define:

\(\displaystyle C=c_2-c_1\)

\(\displaystyle F(y)=G(x)+C\)

You can think of both constants being contained in the one, which is generally put on the RHS.

Does that make sense?
 
  • #7
karush said:
$\tiny{2.3.2}$
$\textsf{Given: }$
$$\displaystyle y'=2xy^2, \quad y(0)=1$$
$\textit{Show that $y=\phi(x)=(1-x^2)^{-1}$ is a solution of the initial value problem}
The problem, as stated, does NOT ask you to solve the equation, just to show that the given function is a solution.

With $y= (1- x^2)^{-1}$, $y'= (-1)(1- x^2)^{-2}(-2x)$
Since $y= (1- x^2)^{-1}$, $(1- x^2)^{-2}= [(1- x^2)^{-1}]^2= y^2$
so $y'= 2xy^2$ and we are done!

$
\begin{align*}\displaystyle
y'&=2xy^2\\
\frac{dy}{y^2}&=2x \\
y&=\color{red}{\frac{1}{(c_1-x^2)}}
\end{align*}
ok I went into confusion after 2x??
red is W|F
 
Last edited:

FAQ: -aux.2.2.02 y=\phi(x)=(1-x^2)^{-1} is a solution IVP

1. What is an IVP?

An IVP, or initial value problem, is a type of mathematical problem that involves finding a function that satisfies a given differential equation and a set of initial conditions. The initial conditions are usually specified as the value of the function and its derivatives at a certain point.

2. What does the equation y=\phi(x)=(1-x^2)^{-1} represent?

The equation y=\phi(x)=(1-x^2)^{-1} represents a solution to a differential equation. In this case, it is a solution to an initial value problem.

3. How is the solution to an IVP determined?

The solution to an IVP is determined by solving the given differential equation and then using the initial conditions to find the specific values of the function at the given point.

4. What does the term "solution" mean in relation to an IVP?

In the context of an IVP, a solution refers to a function that satisfies both the given differential equation and the specified initial conditions.

5. Why is the function y=\phi(x)=(1-x^2)^{-1} a solution to the IVP?

The function y=\phi(x)=(1-x^2)^{-1} satisfies both the given differential equation and the specified initial conditions. This can be verified by plugging the function into the differential equation and checking that it satisfies the equation, as well as evaluating the function at the given point and comparing it to the specified initial condition.

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