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  • Find the general solution of (1 + tan y) (dx - dy) + 2xdy = 0.

Find the general solution of (1 + \tan y) (dx - dy) + 2xdy = 0.

Answers (1)

$$ (1+\tan y)(d x-d y)+2 x d y=0 $$
\Rightarrow \mathrm{dx}-\mathrm{dy}+\tan \mathrm{y} \mathrm{d} \mathrm{x}-\tan \mathrm{y} \mathrm{dy}+2 \mathrm{xdy}=0$

Divide throughout by dy
\\\Rightarrow \frac{d x}{d y}-1+\tan y \frac{d x}{d y}-\tan y+2 x=0$ \\$\Rightarrow(1+\tan y) \frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}-(1+\tan y)+2 \mathrm{x}=0$
Divide by (1+tany)
\\\Rightarrow \frac{d x}{d y}-1+\frac{2 x}{1+\tan y}=0$ \\$\Rightarrow \frac{d x}{d y}+\left(\frac{2}{1+t a n y}\right) x=1$

\frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{2}{1+\operatorname{tany}}\right) \mathrm{x}=1 \quad \frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}$
Compare
We get
\\ P=\frac{2}{1+\tan y}$ \\$Q=1$
This is the linear differential equation with P and Q as functions of x
\\\Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{Pdy}}$ \\$\Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{2}{1+\tan \mathrm{y}} \mathrm{dy}}$
Put tany =\frac{\sin y}{\cos y}$
$$ \Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{2 \cos y}{\sin y+\cos y}} \mathrm{dy} $$
Adding and subtracting siny in the numerator

$$ \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{\cos y+\sin y+\cos y-\sin y}{\sin y+\cos y}} d y \\ \Rightarrow \mathrm{IF}=e^{\int\left(1+\frac{\cos y-\sin y}{\sin y+\cos y}\right) d y} \\ \Rightarrow \mathrm{IF}=e^{\int 1 \mathrm{~d} y+\int \frac{\cos y-\sin y}{\sin y+\cos y} d y} $$
Consider the integral \int \frac{\cos y-\sin y}{\sin y+\cos y} d y$
Let \sin y+\cos y=t$
Differentiate with respect to y
We get
\\\frac{\mathrm{dt}}{\mathrm{dy}}=\operatorname{cosy}-\sin y$ \\$\mathrm{dt}=(\cos \mathrm{y}-\mathrm{sin} \mathrm{y}) \mathrm{d} \mathrm{y}$

\begin{aligned} &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\int \frac{1}{t} d t\\ &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\log t\\ &\text { Resubstitue }\\ &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\log (\sin y+\cos y)\\ &\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}+\log (\sin y+\cos y)} \end{aligned}

\\\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}} \times \mathrm{e}^{\text {log }(\text { siny }+\cos y)}$ \\$\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}}(\mathrm{sin} \mathrm{y}+\cos \mathrm{y})$
The solution of the linear differential equation will be x(\mathrm{IF})=\int \mathrm{Q}(\mathrm{IF}) \mathrm{d} \mathrm{y}+\mathrm{c}$
Substitute values for Q and IF
\\\Rightarrow x e^{y}(\sin y+\cos y)=\int(1) e^{y}(\sin y+\cos y) d y+c$ \\$\Rightarrow x e^{y}(\sin y+\cos y)=\int\left(e^{y} \sin y+e^{y} \cos y\right) d y+c$
Put \mathrm{e}^{\mathrm{y}}$ sin y$=t$ and differentiate with respect to y
We get

$$ \frac{d t}{d y}=e^{y} \sin y+e^{y} \cos y $$
Which means
$$ \mathrm{dt}=\left(\mathrm{e}^{\mathrm{y}} \sin \mathrm{y}+\mathrm{e}^{\mathrm{y}} \cos \mathrm{y}\right) \mathrm{d} \mathrm{y}+\mathrm{c} $$
Hence
$$ \\ \Rightarrow x e^{y}(\sin y+\cos y)=\int d t+c \\ \Rightarrow x e^{y}(\sin y+\cos y)=t+c $$
Substitute t again
x e^{y}(\sin y+\cos y)=e^{y} \sin y+c$
$$ \Rightarrow \mathrm{x}=\frac{\sin \mathrm{y}}{\sin \mathrm{y}+\cos \mathrm{y}}+\frac{\mathrm{c}}{\mathrm{e}^{\mathrm{y}}(\sin \mathrm{y}+\mathrm{cosy})} $$

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