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\[\LARGE{\color {purple} {PART- A}}\]
\[\color {purple} {1:}\ if\ y\ =\ x^2\ +\ 6\ x\ -\ 15\ \color {red} {Find\ \frac{d^2y}{dx^2}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ y\ =\ x^2\ +\ 6\ x\ -\ 15\ \hspace{15cm}\]
\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y)\ =\ \frac{d}{dx}( x^2)\ +\ 6\ \frac{d}{dx}( x) -\ \frac{d}{dx}(15)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 2\ x\ +\ 6\ (1)\ -\ 0\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ 2\ x\ +\ 6\ \hspace{10cm}\]
\[Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(\frac{dy}{dx})\ =\ \frac{d}{dx}( 2\ x\ +\ 6)\ \hspace{10cm}\]
\[\frac{d^2y}{dx^2}\ =\ 2\ \frac{d}{dx}(x)\ +\ \frac{d}{dx}(6)\ \hspace{10cm}\]
\[\frac{d^2y}{dx^2}\ =\ 2\ (1)\ +\ 0\ \hspace{10cm}\]
\[\frac{d^2y}{dx^2}\ =\ 2\ \hspace{10cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[\LARGE{\color {purple} {PART- C}}\]
\[\color {purple} {3:}\ If\ y\ =\ x^2\ cos\ x,\ \color {red} {prove\ that\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y}\ =\ 0\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ y\ =\ x^2\ cos\ x\ \hspace{15cm}\]
\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y)\ =\ \frac{d}{dx}( x^2\ cos\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^2\ \frac{d}{dx}(cos\ x)\ +\ cos\ x\ \frac{d}{dx}(x^2)\ \hspace{10cm}\]
\[y_1\ =\ x^2\ (-\ sin\ x)\ +\ cos\ x\ 2\ x\ \hspace{10cm}\]
\[y_1\ =\ -\ x^2\ sin\ x\ +\ 2\ x\ cos\ x\ \hspace{10cm}\]
\[Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y_1)\ =\ \frac{d}{dx}(-\ x^2\ sin\ x\ +\ 2\ x\ cos\ x)\ \hspace{10cm}\]
\[y_2\ =\ -\ \frac{d}{dx}(x^2\ sin\ x)\ +\ 2\ \frac{d}{dx}(x\ cos\ x)\ \hspace{10cm}\]
\[y_2\ =\ -\ (x^2\ \frac{d}{dx}(sin\ x)\ +\ sin\ x\ \frac{d}{dx}(x^2))\ +\ 2[x\ \frac{d}{dx}(cos\ x)\ +\ cos\ x\ \frac{d}{dx}(x)]\ \hspace{10cm}\]
\[y_2\ =\ -( x^2\ cos\ x\ +\ sin\ x\ (2\ x))\ +\ 2[x\ (-\ sin\ x)\ +\ cos\ x\ (1)]\ \hspace{10cm}\]
\[y_2\ =\ -\ x^2\ cos\ x\ -\ 2\ x\ sin\ x\ -\ 2 x\ sin\ x\ +\ 2\ cos\ x\ \hspace{10cm}\]
\[y_2\ =\ -\ x^2\ cos\ x\ -\ 4\ x\ sin\ x\ +\ 2\ cos\ x\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y\ \hspace{10cm}\]
\[=\ x^2\ (-\ x^2\ cos\ x\ -\ 4\ x\ sin\ x\ +\ 2\ cos\ x)\ -\ 4\ x\ (-\ x^2\ sin\ x\ +\ 2\ x\ cos\ x)\ +\ (x^2\ +\ 6)(x^2\ cos\ x)\ \hspace{10cm}\]
\[=\ -\ x^4\ cos\ x\ -\ 4\ x^3\ sin\ x\ +\ 2\ x^2\ cos\ x\ +\ 4\ x^3\ sin\ x\ -\ 8\ x^2\ cos\ x\ +\ x^4\ cos\ x+\ 6\ x^2\ cos\ x\ \hspace{10cm}\]
\[=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[\color {purple} {4:}\ If\ xy\ =\ a\ e^x\ +\ b\ e^{-x}\ \color {red} {prove\ that\ xy_2\ +\ 2\ y_1\ =\ xy}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ xy\ =\ a\ e^x\ +\ b\ e^{-x}\ —–\ (1)\ \hspace{15cm}\]
\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(xy)\ =\ a\ \frac{d}{dx}( e^x)\ +\ b\ \frac{d}{dx}( e^{-x}) \hspace{10cm}\]
\[x\ \frac{d}{dx}(y)\ +\ y\ \frac{d}{dx}(x)\ =\ a\ e^x\ +\ b\ e^{-x} (-1)\ \hspace{10cm}\]
\[xy_1\ +\ y\ (1)\ =\ a\ e^x\ -\ b\ e^{-x}\ \hspace{10cm}\]
\[xy_1\ +\ y\ =\ a\ e^x\ -\ b\ e^{-x}\ —-\ (2)\ \hspace{10cm}\]
\[Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(x\ y_1)\ +\ \frac{d}{dx}(y)\ =\ a\ \frac{d}{dx}(e^x)\ -\ b\ \frac {d}{dx}(e^{-x})\ \hspace{10cm}\]
\[x\ \frac{d}{dx}(y_1)\ +\ y_1\ \frac{d}{dx}(x) +\ y_1\ =\ a\ e^x\ -\ b\ e^{-x} (-\ 1)\ \hspace{10cm}\]
\[xy_2\ +\ y_1\ (1)\ +\ y_1 =\ a\ e^x\ +\ b\ e^{-x}\ \hspace{10cm}\]
\[xy_2\ +\ 2\ y_1\ =\ xy\ ——-\ Using\ (1)\ \hspace{10cm}\]
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