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Just wondering if anyone could have a look through nd see if I've done this right? When it comes to the general solution & conditon I'm not sure if I've done this right.
$\displaystyle
\begin{gathered}
y' - 2xy = x \hfill \\
y' = 2xy + x \hfill \\
\smallint \frac{{dy}}
{y} = \smallint 3x \hfill \\
\log (y) = \frac{3}
{2}x^2 \hfill \\
y(x) = A\exp \{ \frac{3}
{2}x^2 \} \hfill \\
\end{gathered}
$
satisfies
$\displaystyle
y(0) = 0
$
is the answer y(x)=A??
I think I might have gone wrong somewhere
yes.. on the third lineQuote:
Originally Posted by dankelly07
check your factoring..
it should be $\displaystyle y' = (2y + 1)x$ right?
so..
$\displaystyle
\smallint \frac{{dy}}
{{2y + 1}} = \smallint xdx
$
$\displaystyle
\frac{1}
{2}\log (|2y + 1|) = \frac{{x^2 }}
{\begin{gathered}
2 \hfill \\
\hfill \\
\end{gathered} } + A
$
$\displaystyle
2y = e^{-x^2 } - 1 + A
$
$\displaystyle
y = A + \frac{{e^{-x^2 } - 1}}
{\begin{gathered}
2 \hfill \\
\hfill \\
\end{gathered} }
$
am I close?
$\displaystyle \frac{1}{2} \log (2y + 1) = \frac{x^2}{2} + A \Rightarrow \log (2y+1) = x^2 + A$
if you take the exponential of both sides, it should turn out to be like this..
$\displaystyle 2y+1 = e^{x^2 + A} = Ae^{x^2}$ and not the one have written.. continue now..
thanks for your help..
so
Ae^(x^2)+1/2
with condition A=1??
I want to be sure I can do this..
$\displaystyle 2y + 1 = A e^{x^2}$Quote:
Originally Posted by dankelly07
$\displaystyle \Rightarrow y = Ae^{x^2} {\color{red} - }\frac{1}{2}$.
Substitute the given boundary condition y(0) = 0, that is, y = 0 when x = 0:
$\displaystyle 0 = A - \frac{1}{2} \Rightarrow A = \frac{1}{2}$.
Therefore $\displaystyle y = \frac{1}{2} e^{x^2} - \frac{1}{2}$.
You should confirm this solution by substitution into the DE.
