First Order

Solve: $\frac{d^2y}{dx^2} = 2 + \frac{1}{x}$

As the equation is of first order, integrate the function twice, i.e. $\frac{dy}{dx} = 2x + ln\,x + C$ and $y = x^2 + x\,ln\,x + Cx + K$

$y = x^2 + x\,ln\,x + Cx + K$

$\mathbf{\frac{dx}{dy} = 1 - y}$

$dx = (1 - y)\,dy$

$\therefore\;\;\;\;\;\;x = y - \frac{y^2}{2} + C$

$\mathbf{\frac{dy}{dx}\;=\;a^2\;+\;y^2}$

$dx\;=\;\frac{dy}{a^2\;+\;y^2}$

$x\;=\;\frac{1}{a}\;arctan\,(\frac{y}{a})\;+\;c$

$\mathbf{\frac{dy}{dx}\;=\;\frac{x^2\;-\;1}{y}}$

$y\,dy\;=\;(x^2\;-\;1)\,dx$

$\therefore\;\;\;\;\;\frac{y^2}{2}\;=\;\frac{x^3}{3}\;-\;x\;+\;C$

$\mathbf{\frac{dy}{dx}\;=\;\frac{y}{x^2\;-\;1}}$

$\frac{dy}{y}\;=\;\frac{dx}{x^2\;-\;1}$

$\therefore\;\;\;\;\;ln\,y\;=\;\frac{1}{2}\;ln\;\;\left(\frac{x\;+\;1}{x\;-\;1} \right)}$

$\therefore\;\;\;\;\;y\;=\;C\;\sqrt{\;\left(\frac{x\;+\;1}{x\;-\;1} \right)}$

Separate the Variables

Integrating

$\frac{dy}{dx}=\frac{1}{cos^2 \tfrac{1}{2}x}$

This is the same as $y=\int \frac{1}{\cos^2 \tfrac{1}{2}x}dx$

which we integrate in the normal way to yield $y=2\tan \tfrac{1}{2} x + C$

$y=2\tan \tfrac{1}{2} x + C$

Solve the following equation given that y=0 at x=0 $\frac{dy}{dx}=\sec y$

$\sec y \equiv \frac{1}{\cos y}$ so $\frac{dy}{dx}=\sec y$ becomes $\int \cos y\;dy=\int dx$ $\therefore\;\;\;\;\;\;\;\sin y=x+A$

But when x=0, y=0 and so A=0

$y=\sin^{-1}x$

Solve the following equation: $\frac{r \tan \theta }{a^2-r^2} = d \theta$ when r=0 at $theta=\frac{\pi }{4}$

Rearrange and integrate $\int {\frac{r}{a^2 - r^2}\dr = \int \frac{1}{\tan \theta} d \theta$

$-\tfrac{1}{2}\ln(a^2-r^2)=\ln \sin \theta +\ln C$ or $\ln \left (\frac{1}{\sqrt{a^2-r^2}} \right )=\ln \left | \sin \theta \cdot C \right |$

$\therefore\;\;\;\;\;\;\frac{1}{\sqrt{a^2 - r^2}} = C \sin \theta$ but r=0 when $theta=\frac{\pi }{4}$ $\therefore\;\;\;\;\;\frac{1}{a}=\frac{C}{\sqrt{2}}$ thus $C=\frac{\sqrt{2}}{a}$

Therefore $\frac{1}{(a^2 - r^2)} = \frac{2}{a^2} \sin^2\theta$

$(a^2-r^2)=\frac{a^2}{2} \textup{cosec}^2 \theta$

If a and b are the radii of concentric spherical conductors at potentials of $V_1$ respectively, then V is the potential at a distance r from the centre. Find the value of V if: $\frac{d}{dr}\left(r^2 \frac{dV}{dr} \right)=0}$ and $V=V_1$ at r=a and $V=0$ at r=b

$\frac{d}{dr}\left(r^2\frac{dV}{dr} \right)=0$

$\therefore\;\;\;\;\;\;r^2\frac{dV}{dr} \right)=A$

$\frac{dV}{A}=\frac{dr}{r^2}$

$\therefore\;\;\;\;\;\;\frac{V}{A}=-\frac{1}{r}\;+\;B$

Substituting in the given values for V and r $\frac{V_1}{A}=-\frac{1}{a}+B$ and $\frac{0}{A}=-\frac{1}{b}+B$ $\therefore\;\;\;\;\;\;B=\frac{1}{b}$ $\therefore\;\;\;\;\;\;\frac{V_1}{A}=-\frac{1}{a}+\frac{1}{b}=-\frac{b-a}{ab}$ Thus $A=-\frac{abV_1}{b-a}$

$V=\left(\frac{1}{r}-\frac{1}{b} \right)\left(\frac{abV_1}{b-a} \right)$

$\frac{dy}{dx} + \frac{1}{x}\times y = x^2$

If each side of te equation is multiplied by x the equation becomes:-

$x\frac{dy}{dx} + y = x^3$ i.e $\frac{d}{dx}(xy) = x^3$ Hence integrating $\;x\,y = \frac{1}{4}x^4 + c$

This equation has been solved by using the obvious integrating factor x. It is possible to find a more general solution by using R as and integrating factor.

Consider the following equation : $R\frac{dy}{dx} + R\,P\,y = R\,Q$

By Inspection the left hand side of this equation must reduce to (Ry) $\therefore\;\;\;\;\;\;R\frac{dy}{dx} + RP\,y = \frac{d}{dx}(Ry) = R\frac{dy}{dx} + y\frac{dR}{dx}$ This gives $RPy = y\,\frac{dR}{dx}$

$\therefore\;\;\;\;\;\;P\,dx = \frac{dR}{R}$ Thus $\int P\,dx = \ln\,R$

$\therefore\;\;\;\;\;\;R = e^{\int P\,dx}$

This gives the rule that to solve $\frac{dy}{dx} + Py = Q$ multiply both sides by an integrating factor of:-

$e^{\int P\,dx}$

Hence the Method of solving this type of equation is :

• Reduce the equation into the form $\mathbf{\frac{dy}{dx} + Py = Q}$

• Multiply through by the Integrating Factor:- $\mathbf{e^{\int P\,dx}}$

• The equation becomes :-$\mathbf{\frac{d}{dx}(Ry) = Q}$

$\frac{dy}{dx} + 2xy = 10\,x\,e^{-x^2}$

The Integrating Factor is $\displaystyle e^{\int 2x\,dx$

Thus the equation becomes :-

$\frac{d\left(y\,e^{x^2 }\right)}{dx} = 10\,x$

$\therefore\;\;\;\;\;\;y\,e^{x^2} = 5\,x^2 + Constant$

$2x^2\;\frac{dy}{dx} - yx = 3$

$\frac{dy}{dx} - \frac{y}{2x} = \frac{3}{2x^2}$

Therefore the integrating factor is $\displaystyle e^{\int- \frac{1}{2x}dx$

$= e^{- \frac{1}{2}lnx}} = e^{ln\,x^{-\frac{1}{2}}} = x^{-\frac{1}{2}$

Thus the equation reduces down to :-

$\frac{d}{dx}\left(y\;x^{-\,\frac{1}{2} }\right) = \frac{3}{2x^2}\times x^{-\frac{1}{2}} = \frac{3}{2}\;x^{-\frac{5}{3}}$

Integrating:-

$\therefore\;\;\;\;\;\;y\;x^{-\,\frac{1}{2}} = \frac{3}{2}\;x^{-\,\frac{3}{2}}\times \left(- \frac{2}{3} \right) + K$

$thus\;\;\;\;\;\;y\;= - \frac{1}{x} + K\sqrt{x}$

$\frac{dy}{dx} + 3y = e^{2x}$

Here the Integrating Factor is $\displaystyle e^{3\,x}$

Multiplying through by this

$e^{3\,x}\;\frac{dy}{dx} + 3\;e^{3\,x}y = e^{5\,x}$

$i.e.\;\;\;\;\;\;\frac{d}{dx}\left(y\;e^{3\,x} \right) = e^{5\,x}$

Integrating $y\;e^{3\,x} = \frac{1}{5}\,e^{5\,x} + C$

$\therefore\;\;\;\;\;\;y = \frac{1}{5}\;e^{2x} + C\;e^{- 3x}$

Is the follow function homogeneous: $\frac{dy}{dx} = \frac{x^2 + y^2}{2x^2}$

Substitute for y=vx, $\frac{dy}{dx} = \frac{x^2 + (vx)^2}{2x^2}$ or $\frac{dy}{dx} = \frac{x^2 + v^2 x^2}{2x^2}$ or $\frac{dy}{dx} = \frac{1 + v^2}{2}$

As all the x have cancelled out, the test is satisfied.

Function is homogeous

Is the follow function homogeneous: $\frac{dy}{dx} = \frac{y^3}{x^2}$

Substitute y=vx,

$\frac{dy}{dx} = \frac{vx^3}{x^2}$ or $\frac{dy}{dx} = xv^3$

so not all the x have been removed from this solution, so the test has failed.

Function is not homogeneous