SUCCESSIVE DIFFERENTIATION (Revision)

\[1.\ If\ xy\ =\ a\ e^x\ +\ b\ e^{-\ x}\ ,\ prove\ that\ xy_2\ +\ 2\ y_1\ =\ xy\ \hspace{15cm}\]

\[\color {black}{Solution:}\ xy\ =\ a\ e^x\ +\ b\ e^{-\ x}\ —–\ (1)\ \hspace{15cm}\]

\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]

\[\frac{d}{dx}(x\ y)\ =\ \frac{d}{dx}( a\ e^x\ +\ b\ e^{-\ x})\ \hspace{10cm}\]

\[x\ \frac{d}{dx}\ (y)\ +\ y\ \frac{d}{dx}\ (x)\ =\ a\ \frac{d}{dx}\ (e^x)\ +\ b\ \frac{d}{dx}(e^{-\ x})\ \hspace{10cm}\]

\[x\ \frac{dy}{dx}\ +\ y(1)\ =\ a\ e^x\ +\ b\ (e^{-\ x})\ (-\ 1)\ \hspace{10cm}\]

\[x\ y_1\ +\ y\ =\ a\ e^x\ -\ b\ e^{-\ x}\ \hspace{10cm}\]

\[Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]

\[\frac{d}{dx}(x\ y_1\ +\ y)\ =\ \frac{d}{dx}(a\ e^x\ -\ b\ e^{-\ x})\ \hspace{10cm}\]

\[x\ \frac{d}{dx}\ (y_1)\ +\ y_1\ \frac{d}{dx}\ (x)\ +\ \frac{d}{dx}\ (y)\ =\ a\ \frac{d}{dx}\ (e^x)\ -\ b\ \frac{d}{dx}(e^{-\ x})\ \hspace{10cm}\]

\[x\ \frac{dy}{dx}\ +\ y_1(1)\ +\ \frac{dy}{dx}\ =\ a\ e^x\ -\ b\ (e^{-\ x})\ (-\ 1)\ \hspace{10cm}\]

\[x\ y_2\ +\ y_1\ +\ y_1\ =\ a\ e^x\ +\ b\ e^{-\ x}\ \hspace{10cm}\]

\[x\ y_2\ +\ 2\ y_1\ =\ xy\ \hspace{2cm}\ using\ (1)\ \hspace{5cm}\]

\[2.\ \ If\ y\ =\ x^2\ Sin\ x,\ prove\ that\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y\ =\ 0\ \hspace{15cm}\]

\[\color {black}{Solution:}\ Refer\ Q.\ No. 6\ solution\ in\ https://yanamtakshashila.com/2021/11/23/successive-differentiation/ \hspace{10cm}\]