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 \[\dfrac{1}{2}\left(y^2\right)^{\prime} + y^{\prime\prime}=0\] \[\left(\dfrac{1}{2}y^2 + y^{\prime}\right)^{\prime}=0\] \[\dfrac{d}{dx}\left(\dfrac{1}{2}y^2 + y^{\prime}\right)=0\] \[\int \dfrac{d}{dx}\left(\dfrac{1}{2}y^2 + y^{\prime}\right)dx=\int 0\cdot dx\] \[\int d\left(\dfrac{1}{2}y^2 + y^{\prime}\right)=\int 0\cdot dx\] \[\dfrac{1}{2}y^2 + y^{\prime}=C\] on \(C\) és una constant que agrupa les contants de les dues integrals. 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 \[ \dfrac{2}{2C - y^2} = \dfrac{1}{\sqrt{2C}}\left(\dfrac{1}{\sqrt{2C}+y} + \dfrac{1}{\sqrt{2C}-y}\right) \] \[ \dfrac{2}{2C - y^2}dy =dx\] \[ \dfrac{1}{\sqrt{2C}}\left(\dfrac{1}{\sqrt{2C}+y} + \dfrac{1}{\sqrt{2C}-y}\right)dy =dx\] \[ \int\dfrac{1}{\sqrt{2C}+y}dy + \int\dfrac{1}{\sqrt{2C}-y}dy =\int\sqrt{2C}dx\] \[ \ln|\sqrt{2C}+y| - \ln|\sqrt{2C}-y| =\sqrt{2C}x+C'\] \[ \ln\left|\dfrac{\sqrt{2C}+y}{\sqrt{2C}-y}\right| =\sqrt{2C}x+C'\] \[ \dfrac{\sqrt{2C}+y}{\sqrt{2C}-y} =e^{\sqrt{2C}x+C'} =e^{C'}\cdot e^{\sqrt{2C}x} =k\cdot e^{\sqrt{2C}x}\] \[ \sqrt{2C}+y =\sqrt{2C}k\cdot e^{\sqrt{2C}x}-k\cdot e^{\sqrt{2C}x}y\] \[ y +k\cdot e^{\sqrt{2C}x}y=\sqrt{2C}k\cdot e^{\sqrt{2C}x}-\sqrt{2C}\] \[ y\left(1+k\cdot e^{\sqrt{2C}x}\right) y=\sqrt{2C}\left(k\cdot e^{\sqrt{2C}x}-1\right)\] \[ y = \sqrt{2C}\dfrac{k\cdot e^{\sqrt{2C}x}-1}{k\cdot e^{\sqrt{2C}x}-1}\] \[ y = \sqrt{2C}\dfrac{e^{\sqrt{2C}x}-\dfrac{1}{k}}{e^{\sqrt{2C}x}+\dfrac{1}{k}}\] 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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