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[SOLVED] 2.3.2 Show that y=\phi(x)=(1-x^2)^{-1} is a solution of the initial value problem

karush

Well-known member
Jan 31, 2012
2,928
$\tiny{2.3.2}$
$\textsf{Given: }$
$$\displaystyle y'=2xy^2, \quad y(0)=1$$
$\textit{Show that $y=\phi(x)=(1-x^2)^{-1}$ is a solution of the initial value problem}$
\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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

karush

Well-known member
Jan 31, 2012
2,928
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 $\phi(x)$ ??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

karush

Well-known member
Jan 31, 2012
2,928
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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
756
$\tiny{2.3.2}$
$\textsf{Given: }$
$$\displaystyle y'=2xy^2, \quad y(0)=1$$
$\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
 
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