Differential Equations

$ydx + \left( {x + {x^2}y} \right)dy = 0$

$ \Rightarrow {{dx} \over {dy}} = - {x \over y} - {x^2}$

$ \Rightarrow {{dx} \over {dy}} + {x \over y} = - {x^2},$

It is Bernoullis form. Divide by ${x^2}$

${x^{ - 2}}{{dx} \over {dy}} + {x^{ - 1}}\left( {{1 \over y}} \right) = - 1.$

put ${x^{ - 1}} = t,\,\, - {x^{ - 2}}{{dx} \over {dy}} = {{dt} \over {dy}}$

We get, $ - {{dt} \over {dy}} + t\left( {{1 \over y}} \right) = - 1$

$ \Rightarrow {{dt} \over {dy}} - \left( {{1 \over y}} \right)t = 1$

It is linear in $t.$

Integrating factor

$ = {e^{\int { - {1 \over y}dy} }} = {e^{ - \log y}} = {y^{ - 1}}$

$\therefore$ Solution is

$t\left( {{y^{ - 1}}} \right) = \int {\left( {{y^{ - 1}}} \right)} dy + c$

$ \Rightarrow {1 \over x}.{1 \over y} = \log y + c$

$ \Rightarrow \log y - {1 \over {xy}} = c$

General equation of parabola whose axis is along X-axis

${y^2} = 4a\left( {x - h} \right)$

Differentiating with respect to x, we get

$2y{y_1} = 4a$

$ \Rightarrow y{y_1} = 2a$

Again differentiating with respect to x, we get

$ \Rightarrow {y_1}^2 + y{y_2} = 0$

Degree $=1,$ order $=2.$

$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0$

$ \Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}$

$I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } = {e^{{{\tan }^{ - 1}}y}}$

$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \int {{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}}} {e^{{{\tan }^{ - 1}}y}}\,dy$

$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {{{e^{2{{\tan }^{ - 1}}y}}} \over 2} + C$

$\therefore$ $\,\,\,\,\,2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$

${\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} =$ (4)³${\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$

$ \Rightarrow {\left( {1 + 3{{d\,y} \over {d\,x}}} \right)^2} = 16{\left( {{{{d^3}y} \over {d\,{x^3}}}} \right)^3}$

$ \therefore $ Order = 3 and Degree = 3

${{{a^2}y} \over {d{x^2}}} = {e^{ - 2x}};{\mkern 1mu} $

${{dy} \over {dx}} = {{{e^{ - 2x}}} \over { - 2}} + c;$

$y = {{{e^{ - 2x}}} \over 4} + cx + d$