Colector Solar Ejercicio Resuelto

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\begin{equation*} x\frac{dy}{dx} =y+\sqrt{x^{2} +y^{2}} \ \ i\ y( 1) =\sin h\left(\frac{3}{2}\right) \end{equation*}
\begin{gather*} \Longrightarrow \ \frac{dy}{dx} \ =\ \frac{y}{x} +\frac{\sqrt{x^{2} +y^{2}}}{x}\\ \Longrightarrow \ \frac{dy}{dx} \ =\ \frac{y}{x} +\frac{\sqrt{x^{2} +y^{2}}}{x^{2}}\\ \Longrightarrow \ \frac{dy}{dx} \ =\ \frac{y}{x} +\sqrt{1+\left(\frac{y}{x}\right)^{2}}\\ \end{gather*} if a =
\begin{equation*} \frac{y}{x} \end{equation*} then:
\begin{gather*} y=ux\ \\ \Longrightarrow \ \frac{dy}{dx} =u+x\frac{du}{dx} \end{gather*} so:
\begin{gather*} \Longrightarrow \ \frac{dy}{dx} \ =\ \frac{y}{x} +\sqrt{1+(\frac{y}{x}})^{2}\\ \\ \Longrightarrow \ u+x\frac{du}{dx} =u+\sqrt{1+a^{2}}\\ \end{gather*} Thus:
\begin{gather*} x\frac{du}{dx} =u+\sqrt{1+u^{2}} -u\\ \\ \Longrightarrow \ x\frac{du}{dx} =\sqrt{1+u^{2}}\\ \\ \Longrightarrow \ \frac{du}{\sqrt{1+u^{2}}} =\frac{dx}{x} \end{gather*}
\begin{gather*} \frac{du}{\sqrt{1+u^{2}} \ =\frac{dx}{x}}\\ \\ \Longrightarrow \int \frac{du}{\sqrt{1+u^{2}}} =\int \frac{dx}{x} +c\\ \\ \Longrightarrow \ a\mid c\sin h( u) =\mathscr{L} n\mid x\mid +c\\ \Longrightarrow \ u\ =\sin h(\mathscr{L} n\mid x\mid +c) \end{gather*} By formula:
\begin{equation*} \int \frac{dz}{\sqrt{1+Z^{2}}} \end{equation*}
\begin{equation*} =\ a( c\sin h( z) +C \end{equation*} Going back to:
\begin{equation*} u=\frac{y}{x} 1 \end{equation*} We have:
\begin{gather*} \frac{y}{x} =\sin h(\mathscr{L} n\mid x\mid +c)\\ y=x\sin h(\mathscr{L} n\mid x\mid +c) \end{gather*} Now, since
\begin{gather*} y( 1) =\sin h\left(\frac{3}{2}\right)\\ \Longrightarrow \ \sin h\left(\frac{3}{2}\right) =( 1)\sin h(\mathscr{L} n\mid 1\mid +c)\\ \Longrightarrow \ \sin h\left(\frac{3}{2}\right) =\sin h(\mathscr{L} n\mid 1\mid +c)\\ \Longrightarrow \frac{3}{2} =\mathscr{L} n\mid 1\mid +c\\ \frac{3}{2} =0+c\\ c=\frac{3}{2}\\ \end{gather*} Therefore
\begin{equation*} y=x\sin h\left(\mathscr{L} n\mid x\mid +\frac{3}{2}\right) \end{equation*}

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