# SUCCESSIVE DIFFERENTIATION (Excercise)

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1.}\ \color {red} {Find\ the\ order\ and\ degree\ of\ the\ differential\ equation}\ \frac{d^3y}{dx^3}\ -\ 5\ \frac{d^2y}{dx^2}\ +\ 6\ \frac{dy}{dx}\ +\ 7\ y\ =\ 0\ \hspace{10cm}$
$\color {blue}{Solution:}\ order\ =\ 3,\ degree\ =\ 1\ \hspace{15cm}$
$\color {purple} {2:}\ if\ y\ =\ x^2\ +\ 6\ x\ -\ 15\ \color {red} {Find\ \frac{d^2y}{dx^2}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ x^2\ +\ 6\ x\ -\ 15\ \hspace{15cm}$
$Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}(y)\ =\ \frac{d}{dx}( x^2)\ +\ 6\ \frac{d}{dx}( x) -\ \frac{d}{dx}(15)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 2\ x\ +\ 6\ (1)\ -\ 0\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ 2\ x\ +\ 6\ \hspace{10cm}$
$Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}(\frac{dy}{dx})\ =\ \frac{d}{dx}( 2\ x\ +\ 6)\ \hspace{10cm}$
$\frac{d^2y}{dx^2}\ =\ 2\ \frac{d}{dx}(x)\ +\ \frac{d}{dx}(6)\ \hspace{10cm}$
$\frac{d^2y}{dx^2}\ =\ 2\ (1)\ +\ 0\ \hspace{10cm}$
$\frac{d^2y}{dx^2}\ =\ 2\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {3.}\ \color {red} {Form\ the\ differential\ equation\ of}\ y^2\ =\ 4\ a\ x\ by\ eliminating\ the\ constant\ ‘a’\ \hspace{15cm}$
$\color {blue}{Solution:}\ y^2\ =\ 4\ a\ x\ \hspace{15cm}$
$Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}( y^2)\ =\ 4\ a\ \frac{d}{dx}( x)\ \hspace{10cm}$
$2\ y\ \frac{d}{dx}(y)\ =\ 4\ a\ (1)\ \hspace{10cm}$
$2\ y\ \frac{dy}{dx}\ =\ 4\ a\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{4\ a}{2\ y}\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ \frac{2\ a}{y}\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {4:}\ If\ y\ =\ x^2\ cos\ x,\ \color {red} {prove\ that\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y}\ =\ 0\ \hspace{15cm}$
$\color {blue}{Solution:}\ y\ =\ x^2\ cos\ x\ \hspace{15cm}$
$Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}(y)\ =\ \frac{d}{dx}( x^2\ cos\ x)\ \hspace{10cm}$
$\frac{dy}{dx}\ =\ x^2\ \frac{d}{dx}(cos\ x)\ +\ cos\ x\ \frac{d}{dx}(x^2)\ \hspace{10cm}$
$y_1\ =\ x^2\ (-\ sin\ x)\ +\ cos\ x\ 2\ x\ \hspace{10cm}$
$y_1\ =\ -\ x^2\ sin\ x\ +\ 2\ x\ cos\ x\ \hspace{10cm}$
$Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}(y_1)\ =\ \frac{d}{dx}(-\ x^2\ sin\ x\ +\ 2\ x\ cos\ x)\ \hspace{10cm}$
$y_2\ =\ -\ \frac{d}{dx}(x^2\ sin\ x)\ +\ 2\ \frac{d}{dx}(x\ cos\ x)\ \hspace{10cm}$
$y_2\ =\ -\ (x^2\ \frac{d}{dx}(sin\ x)\ +\ sin\ x\ \frac{d}{dx}(x^2))\ +\ 2[x\ \frac{d}{dx}(cos\ x)\ +\ cos\ x\ \frac{d}{dx}(x)]\ \hspace{10cm}$
$y_2\ =\ -( x^2\ cos\ x\ +\ sin\ x\ (2\ x))\ +\ 2[x\ (-\ sin\ x)\ +\ cos\ x\ (1)]\ \hspace{10cm}$
$y_2\ =\ -\ x^2\ cos\ x\ -\ 2\ x\ sin\ x\ -\ 2 x\ sin\ x\ +\ 2\ cos\ x\ \hspace{10cm}$
$y_2\ =\ -\ x^2\ cos\ x\ -\ 4\ x\ sin\ x\ +\ 2\ cos\ x\ \hspace{10cm}$
$L.\ H.\ S\ =\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y\ \hspace{10cm}$
$=\ x^2\ (-\ x^2\ cos\ x\ -\ 4\ x\ sin\ x\ +\ 2\ cos\ x)\ -\ 4\ x\ (-\ x^2\ sin\ x\ +\ 2\ x\ cos\ x)\ +\ (x^2\ +\ 6)(x^2\ cos\ x)\ \hspace{10cm}$
$=\ -\ x^4\ cos\ x\ -\ 4\ x^3\ sin\ x\ +\ 2\ x^2\ cos\ x\ +\ 4\ x^3\ sin\ x\ -\ 8\ x^2\ cos\ x\ +\ x^4\ cos\ x+\ 6\ x^2\ cos\ x\ \hspace{10cm}$
$=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}$
$\color {purple} {5:}\ If\ xy\ =\ a\ e^x\ +\ b\ e^{-x}\ \color {red} {prove\ that\ xy_2\ +\ 2\ y_1\ =\ xy}\ \hspace{15cm}$
$\color {blue}{Solution:}\ xy\ =\ a\ e^x\ +\ b\ e^{-x}\ —–\ (1)\ \hspace{15cm}$
$Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}(xy)\ =\ a\ \frac{d}{dx}( e^x)\ +\ b\ \frac{d}{dx}( e^{-x}) \hspace{10cm}$
$x\ \frac{d}{dx}(y)\ +\ y\ \frac{d}{dx}(x)\ =\ a\ e^x\ +\ b\ e^{-x} (-1)\ \hspace{10cm}$
$xy_1\ +\ y\ (1)\ =\ a\ e^x\ -\ b\ e^{-x}\ \hspace{10cm}$
$xy_1\ +\ y\ =\ a\ e^x\ -\ b\ e^{-x}\ —-\ (2)\ \hspace{10cm}$
$Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}$
$\frac{d}{dx}(x\ y_1)\ +\ \frac{d}{dx}(y)\ =\ a\ \frac{d}{dx}(e^x)\ -\ b\ \frac {d}{dx}(e^{-x})\ \hspace{10cm}$
$x\ \frac{d}{dx}(y_1)\ +\ y_1\ \frac{d}{dx}(x) +\ y_1\ =\ a\ e^x\ -\ b\ e^{-x} (-\ 1)\ \hspace{10cm}$
$xy_2\ +\ y_1\ (1)\ +\ y_1 =\ a\ e^x\ +\ b\ e^{-x}\ \hspace{10cm}$
$xy_2\ +\ 2\ y_1\ =\ xy\ ——-\ Using\ (1)\ \hspace{10cm}$