First Order Differential Equation

e? (dy/dx) - 2e? sinx + sinxcos²x = 0
d/dx (e? ) - (2sinx)e? = -sinxcos²x
I.F. = e^ (-∫2sinxdx) = e²cosx
Solution: e? e²cosx = -∫e²cosx sinx cos²x dx
Let cosx=t, -sinxdx=dt
∫e²? t²dt = e²? t²/2 - ∫2te²? /2 dt = e²? t²/2 - [te²? /2 - ∫e²? /2 dt] = e²? t²/2 - te²? /2 + e²? /4 + C
e^ (y+2cosx) = e²cosxcos²x/2 - e²cosxcosx/2 + e²cosx/4 + C
y (π/2)=0 ⇒ e? = 0 + 0 + e? /4 + C ⇒ C=3/4
y = ln (e²cosx (cos²x/2-cosx/2+1/4)+3/4e? ²cosx)
y (0) = ln (e² (1/2-1/2+1/4)+3/4e? ²) = ln (e²/4+3/4e? ²)
This seems very complex. The solution provided leads to α=1/4, β=3/4. 4 (α+β)=4.

$\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y}\Rightarrow {\displaystyle \int \frac{2^{y}dy}{2^y-1}={\displaystyle \int 2^{x}dx\Rightarrow put{2}^y-1=t\Rightarrow 2^{y}ln2dy=dt}}$

$\frac{1}{ln2}{\displaystyle \int \frac{dt}{t}={\displaystyle \int 2^{x}dx\Rightarrow \frac{1}{ln2}lnt=\frac{2^x}{ln2}+\frac{C}{ln2}}}$    put x = 0 and y = 1 we get C = -1

$\therefore ln\left(2^y-1\right)=2^x-1$    put x = 1 and we get y = log₂ (1 + e)

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

Let x – 1 = X, y – 1 = Y

then DE: $\frac{dY}{dX}=\frac{X+Y}{X-Y}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$

Put y = vx

then $\frac{dY}{dX}=V+X\frac{dV}{dX}$

V + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

$X\frac{dV}{dX}=\frac{1+V}{1-V}-V$

$=\frac{1+V^2}{1-V}$

$\frac{1-V}{1+V^2}dV=\frac{dX}{X}$

$\frac{V-1}{V^2+1}dV+\frac{dX}{X}=0$

$\frac{1}{2}\mathcal{l}n\left|V^2+1\right|-ta{n}^{-1}V+\mathcal{l}n\left|X\right|=c$

$\mathcal{l}n\sqrt[]{V^2+1}\cdot X-ta{n}^{-1}V=c$

$\mathcal{l}n\left(\sqrt[]{1+{\left(\frac{Y-1}{X-}\right)}^2}\cdot \left|X-1\right|\right)ta{n}^{-1}\frac{Y-1}{X-1}=c$

$\left(2,1\right)\Rightarrow \mathcal{l}n\left(\sqrt[]{1+0}\cdot 1\right)-0=c$

c = 0

$\mathcal{l}n\sqrt[]{{\left(X-1\right)}^2+{\left(Y-1\right)}^2}=ta{n}^{-1}\frac{Y-1}{X-1}$

point (k + 1, 2) $\mathcal{l}n\sqrt[]{k^2+1}=ta{n}^{-1}\frac{1}{k}$

$\Rightarrow \frac{1}{2}\mathcal{l}n\left(k^2+1\right)=ta{n}^{-1}\frac{1}{k}$

$sin\left(2x^2\right).109_e\left(tan{x}^2\right)dy+4xydx=4\sqrt[]{2}.x.\left(sin{x}^{2}cos\frac{\pi }{4}-cos{x}^{2}sin\frac{\pi }{4}\right)dx$

$\Rightarrow \mathcal{l}n\left(tan{x}^2\right)dy+\frac{4x}{sin\left(2x^2\right)}ydx=\frac{4x\left(sin{x}^2-cos{x}^2\right)}{sin\left(2x^2\right)}dx$

Integrate

$\Rightarrow y.\mathcal{l}n\left(tan{x}^2\right)=2.\mathcal{l}n\left(\frac{sin{x}^2+cos{x}^2-1}{sin{x}^2+cos{x}^2+1}\right)+C$

$x=\sqrt[]{\frac{\pi }{6}},y=1$ calculate C.

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

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

$Letxy=v\Rightarrow x\frac{dy}{dx}+y=\frac{dv}{dx}$

Put x = 1, y = 1 $\Rightarrow ta{n}^{-1}=c\Rightarrow c=\frac{\pi }{4}$

$\therefore ta{n}^{-1}\left(xy\right)=\mathcal{l}nx=\frac{\pi }{4}$

$e\left(y\left(e\right)\right)=tan\left(1+\frac{\pi }{4}\right)=\frac{tan1+1}{1-tan1}$