MHB -1.5.7.9 find gereral solution to DE

  • Thread starter Thread starter karush
  • Start date Start date
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\tiny{1.5.7.9}$

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
 
Physics news on Phys.org
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 $C'= e^C$.
 
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*} $
 
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*} $
 
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
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top