-1.5.7.9 find gereral solution to DE

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  • Thread starter karush
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In summary, to find the general solution for $y'+\dfrac{3}{x^2}y=0$, you can use an integrating factor of $e^{\frac{3}{x}}$. This results in the solution $y=Ce^{\frac{3}{x}}$, where $C$ is a constant. The constant can be rewritten as $C'=e^C$, giving the final solution of $y=C'e^{\frac{3}{x}}$.
  • #1
karush
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$\tiny{1.5.7.9}$

Find the general solution
$y'+\dfrac{3}{x^2}y=0$

$ u(x)=e^{\displaystyle\int \dfrac{3}{x^2} dx}=e^{-\dfrac{3}{x}}+c$

just seeing if this is started ok... not sure if the given eq should have been rewritten
 
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  • #2
You have misplaced the constant, c. It should also be in the exponential.

$\frac{dy}{dx}= -\frac{3}{x^2}y$
$\frac{dy}{y}= -\frac{3}{x^2}dx= -3x^{-2}dx$
Integrate both sides
$ln(y)= 3x^{-1}+ C= \frac{3}{x}+ C$
Take the exponential of both sides
$y= e^{\frac{3}{x}+ C}= e^Ce^{\frac{3}{x}}= C' e^{\frac{3}{x}}$
where $C'= e^C$.
 
  • #3
karush said:
$\tiny{1.5.7.9}$

Find the general solution
$y'+\dfrac{3}{x^2}y=0$

$ u(x)=e^{\displaystyle\int \dfrac{3}{x^2} dx}=e^{-\dfrac{3}{x}}+c$

just seeing if this is started ok... not sure if the given eq should have been rewritten

Since you are trying to use an integrating factor...

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{3}{x}}\,y' + \frac{3}{x^2}\,\mathrm{e}^{-\frac{3}{x}}\,y &= 0 \\
\left( \mathrm{e}^{-\frac{3}{x}}\,y \right) ' &= 0 \\
\mathrm{e}^{-\frac{3}{x}}\,y &= \int{0\,\mathrm{d}x} \\
\mathrm{e}^{-\frac{3}{x}}\,y &= C \\
y &= C\,\mathrm{e}^{\frac{3}{x}} \end{align*} $
 
  • #4
karush said:
$\tiny{1.5.7.9}$

Find the general solution
$y'+\dfrac{3}{x^2}y=0$

$ u(x)=e^{\displaystyle\int \dfrac{3}{x^2} dx}=e^{-\dfrac{3}{x}}+c$

just seeing if this is started ok... not sure if the given eq should have been rewritten

Since you are trying to use an integrating factor...

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{3}{x}}\,y' + \frac{3}{x^2}\,\mathrm{e}^{-\frac{3}{x}}\,y &= 0 \\
\left( \mathrm{e}^{-\frac{3}{x}}\,y \right) ' &= 0 \\
\mathrm{e}^{-\frac{3}{x}}\,y &= \int{0\,\mathrm{d}x} \\
\mathrm{e}^{-\frac{3}{x}}\,y &= C \\
y &= C\,\mathrm{e}^{\frac{3}{x}} \end{align*} $
 
  • #5
Prove It said:
Since you are trying to use an integrating factor...

$\displaystyle \begin{align*} \mathrm{e}^{-\frac{3}{x}}\,y' + \frac{3}{x^2}\,\mathrm{e}^{-\frac{3}{x}}\,y &= 0 \\
\left( \mathrm{e}^{-\frac{3}{x}}\,y \right) ' &= 0 \\
\mathrm{e}^{-\frac{3}{x}}\,y &= \int{0\,\mathrm{d}x} \\
\mathrm{e}^{-\frac{3}{x}}\,y &= C \\
y &= C\,\mathrm{e}^{\frac{3}{x}} \end{align*} $
yes,, I came close to this but got the product wrong...:unsure:
Mahalo
 

FAQ: -1.5.7.9 find gereral solution to DE

1. What is a general solution to a differential equation?

A general solution to a differential equation is an equation that satisfies the given differential equation for all possible values of the independent variable.

2. How do you find the general solution to a differential equation?

To find the general solution to a differential equation, you need to integrate the equation and include a constant of integration. This will result in a family of solutions that can be used to solve the differential equation for any given set of initial conditions.

3. What is the role of the constant of integration in the general solution?

The constant of integration represents the unknown constants that are present in the general solution. These constants are determined by the initial conditions of the differential equation.

4. Can a general solution be used to solve a specific differential equation?

Yes, a general solution can be used to solve a specific differential equation by substituting the initial conditions into the equation and solving for the constants of integration. This will result in a unique solution for the specific differential equation.

5. Are there any limitations to finding a general solution to a differential equation?

Yes, there are limitations to finding a general solution to a differential equation. Some differential equations may not have a closed-form general solution, and numerical methods may need to be used to approximate a solution. Additionally, the general solution may not be valid for all values of the independent variable due to restrictions or singularities in the equation.

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