Solve 2.1.5 DE w/ $y(0)=1$ & Find $\mu(x)$

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  • Thread starter karush
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In summary, the initial value y(0) of the equation is 1, and the equation becomes an IVP when y is given the initial value 1. The equation has a solution when y(0)=1. The solution is y(x)=2(x-1)e^{2x}+c_1e^x. The constant of integration is c_1=0.
  • #1
karush
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find y
$$\displaystyle y^\prime - y =2xe^{2x}, \quad y(0)=1$$

obtain u(x)
$$\displaystyle\mu(x)
=\exp\left(\int (1),dx\right)
=e^{x}$$

if ok then proceed to distribute to each term
also, this has $y(0)=1$ so new feature to deal with?
 
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  • #2
Careful...in this case $P(x)=-1$, and so:

\(\displaystyle \mu(x)=\exp\left(-\int\,dx\right)=e^{-x}\)

:)

edit: Being given the initial value $y(0)=1$ turns this into an IVP, which will allow you to determine the constant of integration to get a specific solution that both satisfies the ODE and the given initial condition.
 
  • #3
MarkFL said:
Careful...in this case $P(x)=-1$, and so:

\(\displaystyle \mu(x)=\exp\left(-\int\,dx\right)=e^{-x}\)

:)

edit: Being given the initial value $y(0)=1$ turns this into an IVP, which will allow you to determine the constant of integration to get a specific solution that both satisfies the ODE and the given initial condition.
distribute $e^{-x}$
$$\displaystyle e^{-x}y^\prime - e^{-x}y =2xe^{2x}e^{-x}=2e^x x$$
$$\frac{dy}{dx}(xy)=2e^x x$$
integrate
$$xy=\int 2e^x x \, dx =2e^x(x-1)+c $$
divide by x
$$y=2e^x-\frac{2e^x}{x}+\frac{c}{x}$$
 
Last edited:
  • #4
karush said:
distribute $e^{-x}$
$$\displaystyle e^{-x}y^\prime - e^{-x}y =2xe^{2x}e^{-x}=2e^x x$$
$$\frac{dy}{dx}(xy)=2e^x x$$
integrate
$$xy=\int 2e^x x \, dx =2e^x(x-1)+c $$
divide by x
$$y=2e^x-\frac{2e^x}{x}+\frac{c}{x}$$

What you want is:

\(\displaystyle \frac{d}{dx}\left(e^{-x}y\right)=2xe^x\)

Now you can integrate.

Yous see, after multiplying through by the integrating factor, you should have:

\(\displaystyle \frac{d}{dx}\left(\mu(x)y\right)=\mu(x)Q(x)\)
 
  • #5
$$\displaystyle \frac{d}{dx}\left(e^{-x}y\right)=2xe^x$$
$\textit{integrate } $
$$\displaystyle e^{-x}y=\int 2xe^x dx$$
$$\displaystyle e^{-x}y =2e^x(x-1)+c$$
$\textit{divide by $e^{-x}$}$
$$\displaystyle y=2e^{2x}(x-1)+\frac{c}{e^{-x}}$$
$\textit{so if $y(0)=1$ then}$something goofy!
 
  • #6
I would write the general solution to the ODE as:

\(\displaystyle y(x)=2(x-1)e^{2x}+c_1e^x\)

Now, we can write:

\(\displaystyle y(0)=2(0-1)e^{2\cdot0}+c_1e^0=1\)

Solve this for $c_1$...what do you get?
 
  • #7
\(\displaystyle c_1= 0\)

$\textit{so does this mean $c_1$ can be dropped}$
 
  • #8
karush said:
\(\displaystyle c_1= 0\)

$\textit{so does this mean $c_1$ can be dropped}$

Please check your computation...I get a different result. :)
 

FAQ: Solve 2.1.5 DE w/ $y(0)=1$ & Find $\mu(x)$

What is the meaning of DE in the context of this problem?

DE stands for differential equation. In this context, it refers to a mathematical equation that involves an unknown function and its derivatives.

What does $y(0)=1$ indicate in this problem?

This notation represents an initial condition, where the value of the function y is known at a specific point, in this case when x=0. It is necessary for solving the differential equation.

How do you solve for $y(x)$ in this problem?

To solve for $y(x)$, you need to use a method called separation of variables. This involves separating the equation into two separate equations, one with only x terms and the other with only y terms. Then, you can integrate both sides and solve for y.

What is the role of $\mu(x)$ in this problem?

$\mu(x)$ is known as the integrating factor, and it is used to help solve differential equations that are not in a simple form. It multiplies the entire equation and helps to simplify the solution process.

Is there a specific method for finding $\mu(x)$ or is it arbitrary?

There is a specific method for finding $\mu(x)$, which involves using an integrating factor formula based on the form of the differential equation. It is not arbitrary and requires some mathematical manipulation to determine the correct integrating factor.

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