Troba la solució de \(y\cdot y^\prime + y^{\prime\prime}=0\)
No és una equació lineal però té una solució accessible
Recordem que
\[\left(y^2\right)^{\prime}=2\cdot y\cdot y^{\prime}\]
per tant
\[y\cdot y^{\prime}=\dfrac{1}{2}\left(y^2\right)^{\prime}\]
així l'equació diferencial no lineal es pot escriure com
\[\dfrac{1}{2}\left(y^2\right)^{\prime} + y^{\prime\prime}=0\]
\[\left(\dfrac{1}{2}y^2 + y^{\prime}\right)^{\prime}=0\]
\[\dfrac{d}{dx}\left(\dfrac{1}{2}y^2 + y^{\prime}\right)=0\]
\[\dfrac{1}{2}y^2 + y^{\prime}=C\]
on \(C\) és una constant arbbritraria resultat de la integració. Podem separar les variables
\[
\dfrac{dy}{dx} = y^{\prime}=C - \dfrac{1}{2}y^2
\hspace{50pt}
\dfrac{dy}{C - \dfrac{1}{2}y^2} =dx
\hspace{50pt}
\dfrac{2}{2C - y^2}dy =dx
\]
Simplifiquem el quocient car \(2C - y^2=\left(\sqrt{2C}+y\right)\cdot\left(\sqrt{2C}-y\right)\)
\[ \dfrac{2}{2C - y^2} = \dfrac{A}{\sqrt{2C}+y} + \dfrac{B}{\sqrt{2C}-y} \]
\[ \dfrac{2}{2C - y^2}\cdot\left(2C - y^2\right) = \dfrac{A}{\sqrt{2C}+y}\cdot\left(2C - y^2\right) + \dfrac{B}{\sqrt{2C}-y}\cdot\left(2C - y^2\right) \]
\[ 2 = A\cdot\left(\sqrt{2C}-y\right) + B\cdot\left(\sqrt{2C}+y\right) \]
\[ 2 = A\cdot\sqrt{2C}-A\cdot y + B\cdot\sqrt{2C} + B\cdot y \]
\[ 2 = A\cdot\sqrt{2C} + B\cdot\sqrt{2C} + \left(-A+B\right)\cdot y \]
Així
\[
\left\{
\begin{array}{rl}
2 =& A\cdot\sqrt{2C} + B\cdot\sqrt{2C} \\
0 =& -A + B
\end{array}
\right.
\hspace{6em}
\left\{
\begin{array}{rl}
2 =& 2A\cdot\sqrt{2C}\\
A =& B
\end{array}
\right.
\hspace{6em}
A = B = \dfrac{1}{\sqrt{2C}}
\]
òbviament
\[ \dfrac{2}{2C - y^2} = \dfrac{1}{\sqrt{2C}}\left(\dfrac{1}{\sqrt{2C}+y} + \dfrac{1}{\sqrt{2C}-y}\right) \]
\[ \dfrac{2}{2C - y^2}dy = \dfrac{1}{\sqrt{2C}}\left(\dfrac{1}{\sqrt{2C}+y}dy + \dfrac{1}{\sqrt{2C}-y}\right)dy=dx\]
\[ \int\dfrac{1}{\sqrt{2C}+y}dy + \int\dfrac{1}{\sqrt{2C}-y}dy =\int\sqrt{2C}dx\]
\[ \ln|\sqrt{2C}+y| - \ln|\sqrt{2C}-y| =\sqrt{2C}x+C'\]
\[ \ln\left|\dfrac{\sqrt{2C}+y}{\sqrt{2C}-y}\right| =\sqrt{2C}x+C'\]
\[ \dfrac{\sqrt{2C}+y}{\sqrt{2C}-y} =e^{\sqrt{2C}x+C'} =e^{C'}\cdot e^{\sqrt{2C}x} =k\cdot e^{\sqrt{2C}x}\]
\[ \sqrt{2C}+y =\sqrt{2C}k\cdot e^{\sqrt{2C}x}-k\cdot e^{\sqrt{2C}x}y\]
\[ y +k\cdot e^{\sqrt{2C}x}y=\sqrt{2C}k\cdot e^{\sqrt{2C}x}-\sqrt{2C}\]
\[ y\left(1+k\cdot e^{\sqrt{2C}x}\right) y=\sqrt{2C}\left(k\cdot e^{\sqrt{2C}x}-1\right)\]
\[ y = \sqrt{2C}\dfrac{k\cdot e^{\sqrt{2C}x}-1}{k\cdot e^{\sqrt{2C}x}-1}\]
\[ y = \sqrt{2C}\dfrac{e^{\sqrt{2C}x}-\dfrac{1}{k}}{e^{\sqrt{2C}x}+\dfrac{1}{k}}\]
Com que son constants d'integració arbitrarias podem fer la adjudicació
\[ \sqrt{2C}\rightarrow a \hspace{12em} \dfrac{1}{k} \rightarrow b\]
que simplifica molt la solució
\[
\boxed{y = a\cdot\dfrac{e^{ax}-b}{e^{ax}+b}}
\]
Comprovació
\[
\begin{array}{rl}
y^{\prime} &= a\cdot\dfrac{\left(e^{ax}-b\right)^{\prime}\left(e^{ax}+b\right)-
\left(e^{ax}-b\right)\left(e^{ax}+b\right)^{\prime} }{\left(e^{ax}+b\right)^2} \\
&= a\cdot\dfrac{a\cdot e^{ax}\left(e^{ax}+b\right)-
\left(e^{ax}-b\right)a\cdot e^{ax} }{\left(e^{ax}+b\right)^2}\\
&= a\cdot\dfrac{ a\cdot e^{2ax}+a\cdot b\cdot e^{ax} - a\cdot e^{2ax}+b\cdot a\cdot e^{ax} }{\left(e^{ax}+b\right)^2}\\
&= \dfrac{ 2\cdot a^2\cdot b\cdot e^{ax} }{\left(e^{ax}+b\right)^2} = 2\cdot a^2\cdot b\cdot \dfrac{ e^{ax} }{\left(e^{ax}+b\right)^2}
\end{array}
\]
\[
\begin{array}{rl}
y^{\prime\prime} &= 2\cdot a^2\cdot b\cdot\dfrac{\left(e^{ax}\right)^{\prime}\left(e^{ax}+b\right)^2-
\left(e^{ax}\right)\left(\left(e^{ax}+b\right)^2\right)^{\prime} }{\left(\left(e^{ax}+b\right)^2\right)^2} \\
&= 2\cdot a^2\cdot b\cdot\dfrac{a\cdot e^{ax}\left(e^{ax}+b\right)^2-
e^{ax}2\cdot\left(e^{ax}+b\right)\cdot a\cdot e^{ax}}{\left(e^{ax}+b\right)^4} \\
&= 2\cdot a^2\cdot b\cdot a\cdot e^{ax}\left(e^{ax}+b\right)\dfrac{\left(e^{ax}+b\right)-
2\cdot e^{ax}}{\left(e^{ax}+b\right)^4} \\
&= 2\cdot a^3\cdot b\cdot e^{ax}\dfrac{e^{ax}+b-
2\cdot e^{ax}}{\left(e^{ax}+b\right)^3} \\
&= 2\cdot a^3\cdot b\cdot e^{ax}\dfrac{b- e^{ax}}{\left(e^{ax}+b\right)^3} \\
\end{array}
\]
\[
\begin{array}{rl}
y\cdot y^{\prime} &= a\cdot\dfrac{e^{ax}-b}{e^{ax}+b} \cdot 2\cdot a^2\cdot b\cdot \dfrac{ e^{ax} }{\left(e^{ax}+b\right)^2}\\
&= 2\cdot a^3\cdot b\cdot \dfrac{ e^{ax} \cdot\left(e^{ax}-b\right)}{\left(e^{ax}+b\right)^3}\\
&= 2\cdot a^3\cdot b\cdot e^{ax} \cdot\dfrac{ e^{ax}-b}{\left(e^{ax}+b\right)^3} \\
&= -2\cdot a^3\cdot b\cdot e^{ax} \cdot\dfrac{b- e^{ax}}{\left(e^{ax}+b\right)^3}\\
\end{array}
\]
Òbviament
\[y\cdot y^\prime + y^{\prime\prime}=0\]
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