Differential Equations 3 Question 15
15. Determine the equation of the curve passing through the origin in the form $y=f(x)$, which satisfies the differential equation $\frac{d y}{d x}=\sin (10 x+6 y)$
$(1996,5$ M)
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Solution:
- Given,
$$ \begin{gathered} t _0=\log _{3 / 4}(1 / 2) \\ \frac{d y}{d x}=\sin (10 x+6 y) \end{gathered} $$
Let
$$ \begin{aligned} \Rightarrow & 10+6 \frac{d y}{d x} & =\frac{d t}{d x} \\ \Rightarrow & \frac{d y}{d x} & =\frac{1}{6} \frac{d t}{d x}-10 \end{aligned} $$
Now, the given differential equation becomes
$$ \begin{array}{rlrl} & & \sin t & =\frac{1}{6} \frac{d t}{d x}-10 \\ \Rightarrow & & 6 \sin t & =\frac{d t}{d x}-10 \\ \Rightarrow & \frac{d t}{d x} & =6 \sin t+10 \\ \Rightarrow & \frac{d t}{6 \sin t+10} & =d x \end{array} $$
On integrating both sides, we get
$$ \text { Let } \begin{aligned} \frac{1}{2} \int \frac{d t}{3 \sin t+5}=\int \frac{d t}{3 \sin t+5} & =\int \frac{d t}{3 \frac{2 \tan t / 2}{1+\tan ^{2} t / 2}+5} \\ & =\int \frac{\left(1+\tan ^{2} t / 2\right) d t}{6 \tan \frac{t}{2}+5+5 \tan ^{2} \frac{t}{2}} \end{aligned} $$
Put $\tan t / 2=u$
$\Rightarrow \quad \frac{1}{2} \sec ^{2} t / 2 d t=d u \Rightarrow d t=\frac{2 d u}{\sec ^{2} t / 2}$
$\Rightarrow \quad d t=\frac{2 d u}{1+\tan ^{2} t / 2} \quad \Rightarrow d t=\frac{2 d u}{1+u^{2}}$
$\therefore \quad I _1=\int \frac{2\left(1+u^{2}\right) d u}{\left(1+u^{2}\right)\left(5 u^{2}+6 u+5\right)}=\frac{2}{5} \int \frac{d u}{u^{2}+\frac{6}{5} u+1}$
$=\frac{2}{5} \int \frac{d u}{u^{2}+\frac{6}{5} u+\frac{9}{25}-\frac{9}{25}+1}$
$=\frac{2}{5} \int \frac{d u}{u+\frac{3}{5}^{2}+\frac{4}{5}^{2}}=\frac{2}{5} \cdot \frac{5}{4} \tan ^{-1} \frac{u+3 / 5}{4 / 5}$
$=\frac{1}{2} \tan ^{-1} \frac{5 u+3}{4}=\frac{1}{2} \tan ^{-1} \frac{5 \tan t / 2+3}{4}$
On putting this in Eq. (ii), we get
$$ \begin{aligned} & \frac{1}{4} \tan ^{-1} \frac{5 \tan \frac{t}{2}+3}{4}=x+c \\ \Rightarrow \quad & \tan ^{-1} \frac{5 \tan \frac{t}{2}+3}{4}=4 x+4 c \\ \Rightarrow \quad & \frac{1}{4}[5 \tan (5 x+3 y)+3]=\tan (4 x+4 c) \\ \Rightarrow \quad & 5 \tan (5 x+3 y)+3=4 \tan (4 x+4 c) \end{aligned} $$
When $x=0, y=0$, we get
$5 \tan 0+3=4 \tan (4 c)$
$$ \begin{array}{ll} \Rightarrow & \frac{3}{4}=\tan 4 c \\ \Rightarrow & 4 c=\tan ^{-1} \frac{3}{4} \end{array} $$
Then, $\quad 5 \tan (5 x+3 y)+3=4 \tan 4 x+\tan ^{-1} \frac{3}{4}$
$\Rightarrow \tan (5 x+3 y)=\frac{4}{5} \tan 4 x+\tan \frac{3}{4}-\frac{3}{5}$
$\Rightarrow 5 x+3 y=\tan ^{-1} \frac{4}{5} \tan 4 x+\tan ^{-1} \frac{3}{4} \quad-\frac{3}{5}$
$\Rightarrow \quad 3 y=\tan ^{-1} \frac{4}{5} \tan 4 x+\tan ^{-1} \frac{3}{4} \quad-\frac{3}{5} \quad-5 x$
$\Rightarrow \quad y=\frac{1}{3} \tan ^{-1} \frac{4}{5} \tan 4 x+\tan ^{-1} \frac{3}{4} \quad-\frac{3}{5}-\frac{5 x}{3}$
