242.14.2. solve the initial value problem

[karush]

$\tiny{242.14.2}\\$
$\textsf{(a) Verify that y = $Cx^2+1$ is a general solution to the differential equation $\displaystyle x \frac{dy}{dx}=2y-2$}$
$\textsf{(b) Use part (a) to solve the initial value problem
$\displaystyle x \frac{dy}{dx}=2y-2, \, y(2)=3$}$

$\textit{all new so kinda ??}$

 

[Greg]

$$x\frac{\text{d}y}{\text{d}x}=2y-2\implies\int\frac{1}{2y-2}\,\text{d}y=\int\frac1x\,\text{d}x$$

Can you continue?

 

[MarkFL]

Greg has shown you how to separate variables in order to integrate to obtain the general solution to the given ODE yourself (which I recommend carrying out), but since you are given the solution and asked to confirm it is a solution, you want to take the given solution:

\(\displaystyle y=Cx^2+1\)

And substitute it into the ODE to see if an identity results. In order to do so, we need to compute:

\(\displaystyle \d{y}{x}=2Cx\)

Now, substitute for $y$ and \(\displaystyle \d{y}{x}\) into the ODE...what do you find?

 

[karush]

$$x\frac{\text{d}y}{\text{d}x}=2y-2\implies\int\frac{1}{2y-2}\,\text{d}y=\int\frac1x\,\text{d}x$$

Can you continue?

suggesting that,,,
$$\frac{\ln\left({y-1}\right)}{2}=\ln\left({x}\right)$$

 

[Greg]

Ok good. But we need to include a constant of integration.

$$\frac12\ln|y-1|=\ln|x|+\ln|C|$$

... and now what?

 

[karush]

\(\displaystyle y=Cx^2+1\)

And substitute it into the ODE to see if an identity results. In order to do so, we need to compute:

\(\displaystyle \d{y}{x}=2Cx\)

Now, substitute for $y$ and \(\displaystyle \d{y}{x}\) into the ODE...what do you find?

so
$\displaystyle x \frac{dy}{dx}=2y-2
\implies 2Cx^2=2y-1$
$\text{plug in y?}$

 

[Greg]

so
$\displaystyle x \frac{dy}{dx}=2y-2
\implies 2Cx^2=2y-1$
$\text{plug in y?}$

$2Cx^2=2y-2$

You need to be careful! It's easy to spot the error here but with other problems it may be difficult to do so.

 

[karush]

Ok good. But we need to include a constant of integration.
$$\frac12\ln|y-1|=\ln|x|+\ln|C|$$
... and now what?

if $x=2$ and $y=3$

$$\frac12\ln|y-1|=\ln|x|+\ln|C|$$
$$C=\frac{\sqrt{2}}{2}$$
which means...

 

[Greg]

$$\begin{align*}\frac12\ln|y-1|&=\ln|x|+\ln|C| \\
\ln\sqrt{y-1}&=\ln|Cx| \\
e^{\ln\sqrt{y-1}}&=e^{\ln|Cx|} \\
\sqrt{y-1}&=|Cx| \\
y-1&=Cx^2\quad\text{(just notation here - we calculate }C\text{ using the initial condition).} \\
y&=Cx^2+1\end{align*}$$

$$y(2)=3\implies3=4C+1\implies C=\frac12$$

 

[karush]

confused about the C vs Cx thing?

 

[Greg]

"C vs Cx"? What do you mean?

 

[karush]

nevermind got it

$ln(x)+ln(C)=ln(Cx)$