# -02.2.11 initial value, graph, intervals xdx+ye^{-x}dy=0, \quad y(0)=1

• MHB
• karush
In summary: I couldn't understand it wellIn summary, the solution of the IVP is found by taking the positive root of the equation y^2=-\left(2e^x(x-1)+1\right). The domain of the solution is found by requiring that the left hand side be greater than or equal to zero.
karush
Gold Member
MHB
2000
(a) Find the solution of the given initial value problem in explicit form.
$$xdx+ye^{-x}dy=0, \quad y(0)=1$$
\begin{align*}\displaystyle
xdx&=-ye^{-x}dy \\
\frac{x}{e^{-x}}\, dx&=-y\, dy\\
xe^x\, dx&=-y\, dy
\end{align*}

(b) Plot the graph of the solution
$\quad \textit{ok... I tried some attempts in W|A but my input didn't work}\\$
(c) Determine (at least approximately) the interval in which the solution is defined.
$\quad \textit{...provided by initial value!}$

Last edited:
karush said:
(a) Find the solution of the given initial value problem in explicit form.
$$xdx+ye^{-x}dy=0, \quad y(0)=1$$
\begin{align*}\displaystyle
xdx&=-ye^{-x}dy \\
\frac{x}{e^{-x}}\, dx&=-y\, dy\\
xe^x\, dx&=-y\, dy
\end{align*}

(b) Plot the graph of the solution
$\quad \textit{ok... I tried some attempts in W|A but my input didn't work}\\$
(c) Determine (at least approximately) the interval in which the solution is defined.
$\quad \textit{...provided by initial value!}$

I would arrange the given ODE as:

$$\displaystyle y\,dy=-xe^x\,dx$$

Exchange dummy variables and integrate, using the given boundaries:

$$\displaystyle \int_1^y u\,dy=-\int_0^x ve^v\,dv$$

$$\displaystyle \frac{1}{2}(y^2-1)=-e^x(x-1)-1$$

$$\displaystyle y^2=-\left(2e^x(x-1)+1\right)$$

Given the initial value, we take the positive root:

$$\displaystyle y=\sqrt{-\left(2e^x(x-1)+1\right)}$$

Here's a plot:

View attachment 8665

The domain of the solution is found from:

$$\displaystyle -\left(2e^x(x-1)+1\right)\ge0$$

$$\displaystyle 2e^x(x-1)+1\le0$$

Using numeric techniques, we find (approximately):

$$\displaystyle -1.6783469900166606534\le x\le0.76803904701346556526$$

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total awsome

that was a great help let me see if I can do the next one...

which is.

$\quad\displaystyle \frac{dr}{d\theta}=r^2/\theta \quad r(1)=2$

Ill start a new thread tho
the $\theta$ scares me

wait i don't think i see why the
intregration is set up the way it is,

never saw dummy variables

karush said:
wait i don't think i see why the
intregration is set up the way it is,

never saw dummy variables

Suppose you have the initial value problem (IVP):

$$\displaystyle \frac{dy}{dx}=f(x)$$ where $$\displaystyle y\left(x_0\right)=y_0$$

Now, separating variables and using indefinite integrals, we may write:

$$\displaystyle \int\,dy=\int f(x)\,dx$$

And upon integrating, we find

$$\displaystyle y(x)=F(x)+C$$ where $$\displaystyle \frac{d}{dx}\left(F(x) \right)=f(x)$$

Using the initial condition, we get

$$\displaystyle y\left(x_0 \right)=F\left(x_0 \right)+C$$

Solving for $$C$$ and using $$\displaystyle y\left(x_0\right)=y_0$$, we obtain:

$$\displaystyle C=y_0-F\left(x_0 \right)$$ thus:

$$\displaystyle y(x)=F(x)+y_0-F\left(x_0 \right)$$

which we may rewrite as:

$$\displaystyle y(x)-y_0=F(x)-F\left(x_0 \right)$$

Now, we may rewrite this, using the anti-derivative form of the fundamental theorem of calculus, as:

$$\displaystyle \int_{y_0}^{y(x)}\,dy=\int_{x_0}^{x}f(x)\,dx$$

Now, since the variable of integration gets integrated out, it is therefore considered a "dummy variable" and since it is considered good form not to use the same variable in the boundaries as we use for integration, we may switch these dummy variables and write:

$$\displaystyle \int_{y_0}^{y(x)}\,du=\int_{x_0}^{x}f(v)\,dv$$

This demonstrates that the two methods are equivalent.

Using the boundaries (the initial and final values) in the limits of integration eliminates the need to solve for the constant of integration, and I find it a more intuitive and cleaner approach to separable initial value problems.

well that is pretty cool
most examples I saw had a lot of steps

## 1. What is the initial value in this equation?

The initial value in this equation is the value of y when x is equal to 0. This is denoted by y(0) in the equation.

## 2. How do you graph this equation?

To graph this equation, you can plot a few points by substituting different values for x and solving for y. You can also use a graphing calculator or software to plot the graph for the entire equation.

## 3. What are the intervals in this equation?

The intervals in this equation refer to the range of values that x can take. In this case, there are no specific intervals mentioned, so we can assume that x can take any real value.

## 4. How do you solve this equation?

This equation is a first-order linear differential equation, which can be solved using various methods such as separation of variables, integrating factors, or substitution. The initial value y(0) can be used to find the specific solution to the equation.

## 5. What is the significance of the term e^-x in this equation?

The term e^-x is a mathematical function called the exponential function, which is commonly used to model growth and decay in various natural phenomena. In this equation, it is a coefficient that affects the rate of change of y with respect to x.

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