Question

a) Solve (𝐷

2 + 1)𝑦 = 3𝑥 − 8 cot 𝑥 where 𝐷 =

𝑑

𝑑𝑥

b) Find orthogonal trajectory to the curve given by 𝑟 = 𝑎(1 + cos 𝜃)

Answer 1

Solution:

a):

$(D^2+1)y=3x-8\cot x$

Its auxiliary equation is $m^2+1=0$

$\Rightarrow m=\pm1$

The complementary function is: $y_c(x)=c_1\cos x+c_2\sin x$

Let $u(x)=\cos x,v(x)=\sin x$

Now, particular integral: $y_p(x)=A.u(x)+B.v(x)$

We use the method of parameters.

$A=-\int \dfrac{u(x)Q(x)}{w(u,v)}dx,B=\int \dfrac{v(x)Q(X)}{w(u,v)}dx \\w(u,v)=\begin{vmatrix} u&v\\u'&v'\end{vmatrix}=\begin{vmatrix} \cos x&\sin x\\-\sin x&\cos x\end{vmatrix}=1$

And $Q(x)=3x-8\cot x$

Then, $A=-\int \dfrac{\cos x(3x-8\cot x)}{1}dx$

$=-3\int x\cos xdx+8\int\cos x\cot xdx \\=-3[x \sin x-1\times -\cos x]+8[\ln \tan(x/2)+\cos x] \\=-3x\sin x+5\cos x+8 \ln \tan(x/2)$ [Using integration by parts]

Next, $B=\int \dfrac{\sin x(3x-8\cot x)}{1}dx$

$=3\int x\sin xdx-8\int\sin x\cot xdx \\=3[x\cos x-1\times-\sin x]-8\sin x \\=-3x\cos x-5\sin x$ [Using integration by parts]

So, the particular integral is

$y_p(x)=A.u(x)+B.v(x) \\=(-3x\sin x+5\cos x+8 \ln \tan(x/2))(\cos x)+(-3x\cos x-5\sin x)(\sin x) \\=-6x\sin x \cos x+5[\cos^2x-\sin^2x]+8\ln \tan(x/2) \\=-6x\sin x \cos x+5\cos2x+8\ln \tan(x/2)$

Thus, the solution is

$y=y_c(x)+y_p(x) \\y=c_1\cos x+c_2\sin x-6x\sin x \cos x+5\cos2x+8\ln \tan(x/2)$

b):

Given curve is $r=a(1+\cos \theta)$ ...(i)

Differentiating it, $d r+a \sin \theta d \theta=0$ ...(ii)

On eliminating the parameter ' a ' we get,

$d r+\frac{r \sin \theta}{1+\cos \theta} d \theta=0 \quad or \quad \frac{-1}{r} \frac{d r}{d \theta}=\frac{\sin \theta}{1+\cos \theta}$ ...(iii)

This is the differential equation of the direction field $D_{1}$ for the given family $F_{1}$ .

To find the differential equation of the direction field orthological to $D_{1}$, we replace $\frac{d r}{d \theta}$

by $-r^{2} \frac{d \theta}{d r}$ as evident from the geometry.

Where $\phi_{2}\ and\ \Psi_{2}$ are the angles for the orthogonal trajectory corresponding to the angles $\phi_{1}\ and\ \Psi_{1}$ for the given family (Fig. 1) as shown below.

$\begin{aligned} \text { Clearly, } \quad \phi_{2} &=\left(\phi_{1} \pm \frac{\pi}{2}\right) \text {, whence } \tan \phi_{2}=\tan \left(\phi_{1} \pm \frac{\pi}{2}\right)=-\cot \phi_{1} \\ \left(r \frac{d \theta}{d r}\right)_{2} &=-\left(\frac{1}{r} \frac{d r}{d \theta}\right)_{1} \end{aligned}$

Whence from (3), the differential equation for the orthogonal trajectories becomes

$r \frac{d \theta}{d r}=\frac{\sin \theta}{1+\cos \theta}=\tan \frac{\theta}{2}$

which is a case of variable-separable, and on integration gives

$\log r=2 \log \sin \frac{\theta}{2}+\log 2 c$

This is the equation of the orthogonal family $F_{2}$ .

Since $r=c(1-\cos \theta)$ represents the same curve as $r=c(1+\cos \theta),$ the member of

$F_{2}$ with label c is the same as the member of $F_{1}$ with label $a=-c$ . Thus the given family is selforthogonal.