Answer to Question #228046 in Differential Equations for Aastha

Question

solve the following differential equations

1. (1 + 2 𝑒 𝑥 𝑦) + 2𝑒 𝑥 𝑦 (1 − 𝑥 𝑦 ) 𝑑𝑦 𝑑𝑥 = 0 

2.1. 𝑡𝑎𝑛 𝑥 . 𝑠𝑖𝑛2 𝑦 𝑑𝑥 + 𝑐𝑜𝑠2 𝑥 . 𝑐𝑜𝑡 𝑦 𝑑𝑦 = 0

2. (1 + 2 𝑒 𝑥 𝑦) + 2𝑒 𝑥 𝑦 (1 − 𝑥 𝑦 ) 𝑑𝑦 𝑑𝑥 = 0

3. (𝑥 − 2𝑦 + 1) 𝑑𝑥 + (4𝑥 − 3𝑦 − 6) 𝑑𝑦 = 0

4. 𝑦(𝑥 3 𝑒 𝑥𝑦 − 𝑦) 𝑑𝑥 + 𝑥 (𝑦 + 𝑥 3𝑒 𝑥𝑦) 𝑑𝑦 = 0

5. 𝑦 2 𝑑𝑥 + (𝑥 2 − 𝑥𝑦 − 𝑦 2 ) 𝑑𝑦 = 0

6. (4𝑒 −𝑦 𝑠𝑖𝑛 𝑥 − 1) 𝑑𝑥 − 𝑑𝑦 = 0

7. 𝑥 4 𝑑𝑦 𝑑𝑥 + 𝑥 3𝑦 − 𝑠𝑒𝑐(𝑥𝑦) = 0

8. (1 + 𝑠𝑖𝑛 𝑦) 𝑑𝑥 𝑑𝑦 = 2𝑦 𝑐𝑜𝑠 𝑦 − 𝑥(𝑠𝑒𝑐 𝑦 + 𝑡𝑎𝑛 𝑦) 

Answer 1

QUESTION 1

$$\tan x\cdot\sin^2ydx+\cos^2x\cdot\cot ydy=0\longrightarrow\\[0.3cm] \left.-\frac{\sin x}{\cos x}\cdot\sin^2ydx=\cos^2x\cdot\frac{\cos y}{\sin y}dy\right|\div\left(\sin^2y\cos^2x\right)\\[0.3cm] -\frac{\sin xdx}{\cos^3x}=\frac{\cos ydy}{\sin^3y}\longrightarrow \left[\begin{array}{l} -\sin xdx=d\left(\cos x\right)\\[0.3cm] \cos ydy=d\left(\sin y\right) \end{array}\right]\\[0.3cm] \int\frac{d\left(\cos x\right)}{\cos^3 x}=\int\frac{d\left(\sin y\right)}{\sin^3 y}\longrightarrow\\[0.3cm] \left.-\frac{1}{2\cdot\cos^2x}=-\frac{1}{2\cdot\sin^2y}-\frac{Const}{2}\right|\cdot(-2)\\[0.3cm] \boxed{\frac{1}{\cos^2x}=\frac{1}{\sin^2y}+Const}$$

ANSWER

$$\tan x\cdot\sin^2ydx+\cos^2x\cdot\cot ydy=0\longrightarrow\\[0.3cm] \frac{1}{\cos^2x}=\frac{1}{\sin^2y}+Const-\text{implicit function}$$

QUESTION 6

$$\left(4e^{-y}\sin x-1\right)dx-dy=0\longrightarrow\\[0.3cm] \frac{dy}{dx}=4e^{-y}\sin x-1\longrightarrow\boxed{y(x)=y_{hom.}(x)+y_{nonhom.}(x)}\\[0.3cm] y_{hom.}(x) :\quad\left.\frac{dy}{dx}=4e^{-y}\sin x \right|\cdot\left(dx\cdot e^{y}\right)\\[0.3cm] \int e^{y}dy=\int4\sin xdx\longrightarrow e^{y}=-4\cos x +C_1\longrightarrow\\[0.3cm] \boxed{y_{hom.}(x)=\ln|C_1-4\cos x|}\\[0.3cm] y_{nonhom.}(x) :\quad \frac{dy}{dx}=-1\longrightarrow\boxed{y_{nonhom.}(x)=-x}\\[0.3cm] y(x)=y_{hom.}(x)+y_{nonhom.}(x)=\ln|C_1-4\cos x|-x\\[0.3cm] \boxed{y(x)=\ln|C_1-4\cos x|-x}$$

ANSWER

$$\left(4e^{-y}\sin x-1\right)dx-dy=0\longrightarrow\\[0.3cm] y(x)=\ln|C_1-4\cos x|-x$$

QUESTION 3

"\\left.\\left(x-2y+1\\right)dx+\\left(4x-3y-6\\right)dy=0\\right|\\div\\left(\\left(4x-3y-6\\right)dx\\right)\\\\[0.3cm]\n\\frac{x-2y+1}{4x-3y-6}+\\frac{dy}{dx}=0\\\\[0.3cm]\n\\text{Let}\\quad\\left\\{\\begin{array}{l}\nx=X+a\\longrightarrow dx=dX\\\\[0.3cm]\ny=Y+b\\longrightarrow dy=dY\n\\end{array}\\right.\\quad\\text{, where a,b - const}\\\\[0.3cm]\n\\frac{X+a-2\\left(Y+b\\right)+1}{4\\left(X+a\\right)-3\\left(Y+b\\right)-6}+\\frac{dY}{dX}=0\\\\[0.3cm]\n\\frac{X-2Y+\\left(a-2b+1\\right)}{4X-3Y+\\left(4a-3b-6\\right)}+\\frac{dY}{dX}=0\\\\[0.3cm]\n\\text{Let's choose constants "a" and "b" so}\\quad\\left\\{\\begin{array}{l}\na-2b+1=0\\\\[0.3cm]\n4a-3b-6=0\\end{array}\\right.\\\\[0.3cm]\na-2b+1=4a-3b-6\\longrightarrow3b-2b=4a-a-6-1\\\\[0.3cm]\n\\boxed{b=3a-7}\\longrightarrow\\boxed{a=3\\quad b=2}\\\\[0.3cm]\n\\text{The choice was made to "a" and "b" were intact}\\\\[0.3cm]\n\\text{It remains to solve such an equation} : \\frac{X-2Y}{4X-3Y}+\\frac{dY}{dX}=0\\\\[0.3cm]\n\\text{Let}\\quad Y=VX\\longrightarrow\\frac{dY}{dX}=V+X\\frac{dV}{dX}\\longrightarrow\\\\[0.3cm]\n\\frac{X-2VX}{4X-3VX}+V+X\\frac{dV}{dX}=0\\longrightarrow\\\\[0.3cm]\n-X\\frac{dV}{dX}=V+\\frac{\\cancel{X}\\left(1-2V\\right)}{\\cancel{X}\\left(4-3V\\right)}\\longrightarrow\\\\[0.3cm]\n-X\\frac{dV}{dX}=\\frac{V\\left(4-3V\\right)+1-2V}{4-3V}\\longrightarrow\\\\[0.3cm]\n\\left.-X\\frac{dV}{dX}=\\frac{-3V^2+2V+1}{4-3V}\\right|\\cdot\\left(-\\frac{\\left(4-3V\\right)dX}{X\\left(-3V^2+2V+1\\right)}\\right)\\longrightarrow\\\\[0.3cm]\n\\frac{\\left(4-3V\\right)dV}{3V^2-2V-1}=\\frac{dX}{X}\\\\[0.3cm]\n3V^2-2V-1=0\\longrightarrow\\sqrt{D}=\\sqrt{\\left(-2\\right)^2-4\\cdot3\\cdot\\left(-1\\right)}=4\\\\[0.3cm]\nV_1=\\frac{2+4}{6}=1\\quad\\text{and}\\quad V_2=\\frac{2-4}{6}=-\\frac{1}{3}\\\\[0.3cm]\n3V^2-2V-1=3\\left(V-1\\right)\\left(V+\\frac{1}{3}\\right)\\equiv\\left(V-1\\right)\\left(3V+1\\right)\\\\[0.3cm]\n4-3V=p\\left(V-1\\right)+q\\left(3V+1\\right)\\longrightarrow\\\\[0.3cm]\n4-3V=(p+3q)V+(q-p)\\longrightarrow\\\\[0.3cm]\n\\left\\{\\begin{array}{l}\nq-p=4\\longrightarrow q=p+4\\\\[0.3cm]\np+3(p+4)=-3\\longrightarrow 4p=-15\n\\end{array}\\right.\\longrightarrow\\\\[0.3cm]\n\\boxed{p=-\\frac{15}{4}\\quad\\text{and}\\quad q=\\frac{1}{4}}\\\\[0.3cm]\n\\int\\frac{\\left(-\\displaystyle\\frac{15}{4}\\left(V-1\\right)+\\displaystyle\\frac{1}{4}\\left(3V+1\\right)\\right)dV}{\\left(V-1\\right)\\left(3V+1\\right)}=\\int\\frac{dX}{X}\\\\[0.3cm]\n\\int\\left(-\\frac{15}{4}\\cdot\\frac{1}{3V+1}+\\frac{1}{4}\\cdot\\frac{1}{V-1}\\right)dV=\\int\\frac{dX}{X}\\\\[0.3cm]\n\\left.\\frac{1}{4}\\cdot\\ln|V-1|-\\frac{5}{4}\\cdot\\ln|3V+1|-\\ln|X|=\\ln|C|\\right|\\cdot\\left(-4\\right)\\\\[0.3cm]\n5\\ln|3V+1|+4\\ln|X|-\\ln|V-1|=-4\\ln|C|\\\\[0.3cm]\n\\ln\\left|\\frac{\\left(3V+1\\right)^5X^4}{V-1}\\right|=\\ln\\left|\\frac{1}{C^4}\\right|\\equiv\\ln|c|\\\\[0.3cm]\nY=VX\\longrightarrow V=\\frac{Y}{X}\\longrightarrow\\\\[0.3cm]\n\\frac{\\left(3\\cdot\\displaystyle\\frac{Y}{X}+1\\right)^5X^4}{\\displaystyle\\frac{Y}{X}-1}=c\\longrightarrow\\frac{\\displaystyle\\frac{\\left(3Y+X\\right)^5}{\\cancel{X^5}}\\cdot\\cancel{X^4}}{\\displaystyle\\frac{Y-X}{\\cancel{X}}}=c\\\\[0.3cm]\n\\frac{\\left(3Y+X\\right)^5}{Y-X}=c\\longrightarrow\\left\\{\\begin{array}{l}\nx=X+3\\longrightarrow X=x-3\\\\[0.3cm]\ny=Y+2\\longrightarrow Y=y-2\n\\end{array}\\right.\\\\[0.3cm]\n\\frac{\\left(3\\left(y-2\\right)+x-3\\right)^5}{y-2-\\left(x-3\\right)}=c\\longrightarrow\n\\boxed{\\frac{\\left(3y+x-9\\right)^5}{y-x+1}=c}"

ANSWER

$$\left(x-2y+1\right)dx+\left(4x-3y+6\right)dy=0\longrightarrow\\[0.3cm] \frac{\left(3y+x-9\right)^5}{y-x+1}=c$$

QUESTION 2

$$\left.\left(1-2e^{x/y}\right)dx-2e^{x/y}\cdot\left(1-\frac{x}{y}\right)dy=0\right|\div\left(dy\right)\\[0.3cm] \left(1-2e^{x/y}\right)\cdot\frac{dx}{dy}-2e^{x/y}\cdot\left(1-\frac{x}{y}\right)=0\\[0.3cm] \text{Let}\quad x=yv\longrightarrow\frac{dx}{dy}=v+y\cdot\frac{dv}{dy}\\[0.3cm] \left.\left(1-2e^v\right)\cdot\left(v+y\cdot\frac{dv}{dy}\right)-2e^v\cdot\left(1-v\right)=0\right|\div\left(1-2e^v\right)\\[0.3cm] v+y\cdot\frac{dv}{dy}=\frac{2e^v-2ve^v}{1-2e^v}\longrightarrow y\cdot\frac{dv}{dy}=\frac{2e^v-2ve^v}{1-2e^v}-v\\[0.3cm] y\cdot\frac{dv}{dy}=\frac{2e^v-2ve^v-v+2ve^v}{1-2e^v}\equiv\frac{2e^v-v}{1-2e^v}\\[0.3cm] \left.y\cdot\frac{dv}{dy}=\frac{2e^v-v}{1-2e^v}\right|\cdot\left(\frac{dy}{y}\cdot\frac{1-2e^v}{2e^v-v}\right)\\[0.3cm]\\[0.3cm] \int\frac{1-2e^v}{2e^v-v}dv=\int\frac{dy}{y}\longrightarrow\int\frac{d\left(v-2e^v\right)}{2e^v-v}=\int\frac{dy}{y}\\[0.3cm] -\ln\left|v-2e^v\right|=\ln|y|+\ln|C|\longrightarrow\\[0.3cm] \left.-\ln\left|y\cdot\left(v-2e^v\right)\right|=\ln|C|\right|\cdot\left(-1\right)\\[0.3cm] \ln\left|y\cdot\left(\frac{x}{y}-2e^{x/y}\right)\right|=\ln\left|\frac{1}{C}\right|\equiv\ln|c|\\[0.3cm] \boxed{x-2ye^{x/y}=c}$$

ANSWER

$$\left(1-2e^{x/y}\right)dx-2e^{x/y}\cdot\left(1-\frac{x}{y}\right)dy=0\\[0.3cm] x-2ye^{x/y}=c$$

QUESTION 4

$$𝑦\left(𝑥^3\cdot 𝑒^{𝑥𝑦} − 𝑦\right)𝑑𝑥+𝑥\left(𝑦+𝑥^3\cdot 𝑒^{𝑥𝑦}\right)𝑑𝑦=0$$

Let us check whether this equation in total differentials. Suppose,

$$\underbrace{𝑦\left(𝑥^3\cdot 𝑒^{𝑥𝑦} − 𝑦\right)}_{M(x,y)}𝑑𝑥+\underbrace{𝑥\left(𝑦+𝑥^3\cdot 𝑒^{𝑥𝑦}\right)}_{N(x,y)}𝑑𝑦=0\\[0.3cm] \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}-\text{condition must be fulfilled}\\[0.3cm] \frac{\partial M}{\partial y}=\frac{\partial}{\partial y}\left(𝑦\left(𝑥^3\cdot 𝑒^{𝑥𝑦} − 𝑦\right)\right)= x^3\cdot e^{xy}-y+𝑦\left(𝑥^4\cdot 𝑒^{𝑥𝑦} − 1\right)\\[0.3cm] \boxed{\frac{\partial M}{\partial y}=x^3\cdot e^{xy}-y+𝑦𝑥^4\cdot 𝑒^{𝑥𝑦} − y}\\[0.3cm] \frac{\partial N}{\partial x}=\frac{\partial}{\partial x}\left(x\left(y+𝑥^3\cdot 𝑒^{𝑥𝑦}\right)\right)= y+𝑥^3\cdot 𝑒^{𝑥𝑦}+x\left(3x^2e^{xy}+x^3ye^{xy}\right)\\[0.3cm] \boxed{\frac{\partial N}{\partial x}=y+x^3\cdot e^{xy}+3x^3e^{xy}+yx^4\cdot e^{xy}}\\[0.3cm]$$

As we can see,

$$x^3\cdot e^{xy}+yx^4\cdot e^{xy}-2y\neq y+4x^3\cdot e^{xy}+yx^4\cdot e^{xy}$$

But,

$$\frac{\partial M/\partial y-\partial N/\partial x}{N}=\\[0.3cm] =\frac{x^3\cdot e^{xy}+yx^4\cdot e^{xy}-2y-y-4x^3\cdot e^{xy}-yx^4\cdot e^{xy}}{x\left(y+x^3\cdot e^{xy}\right)}=\\[0.3cm] =\frac{-3\left(y+x^3\cdot e^{xy}\right)}{x\left(y+x^3\cdot e^{xy}\right)}=-\frac{3}{x}-\text{depends only on ''x''}$$

This means that we can find the integrating factor :

$$\int\frac{d\mu}{\mu}=\int\frac{-3dx}{x}\longrightarrow\ln|\mu|=-3\ln|x|\longrightarrow \boxed{\mu=\frac{1}{x^3}}$$

Then,

$$\left.y\left(x^3\cdot e^{xy}-y\right)dx+x\left(y+x^3\cdot e^{xy}\right)dy=0\right|\cdot\left(\frac{1}{x^3}\right)\\[0.3cm] \underbrace{\frac{y\left(x^3\cdot e^{xy}-y\right)}{x^3}}_{F_x(x,y)}\cdot dx+\underbrace{\frac{x\left(y+x^3\cdot e^{xy}\right)}{x^3}}_{F_y(x,y)}\cdot dy=0\\[0.3cm] F_y(x,y)\equiv\frac{\partial F}{\partial y}=\frac{y+x^3\cdot e^{xy}}{x^2}\longrightarrow\\[0.3cm] F(x,y)=\int\frac{y+x^3\cdot e^{xy}}{x^2}\cdot dy+f(x)\longrightarrow\\[0.3cm] F(x,y)=\frac{1}{x^2}\cdot\left(\frac{y^2}{2}+\frac{x^3}{x}\cdot e^{xy}\right)+f(x)\longrightarrow\\[0.3cm] \boxed{F(x,y)=\frac{y^2}{2x^2}+e^{xy}+f(x)}\\[0.3cm] F_x(x,y)\equiv\frac{\partial F}{\partial x}=\frac{\partial}{\partial x}\left(\frac{y^2}{2x^2}+e^{xy}+f(x)\right)=\\[0.3cm] -\frac{y^2}{x^3}+y\cdot e^{xy}+\frac{df}{dx}=\underbrace{\frac{y\left(x^3\cdot e^{xy}-y\right)}{x^3}}_{\text{from the equation}}\\[0.3cm] -\frac{y^2}{x^3}+y\cdot e^{xy}+\frac{df}{dx}=y\cdot e^{xy}-\frac{y^2}{x^3}\longrightarrow\boxed{\frac{df}{dx}=0}\\[0.3cm] \boxed{f(x)=Const}$$

Conclusion,

$$𝑦\left(𝑥^3\cdot 𝑒^{𝑥𝑦} − 𝑦\right)𝑑𝑥+𝑥\left(𝑦+𝑥^3\cdot 𝑒^{𝑥𝑦}\right)𝑑𝑦=0\longrightarrow\\[0.3cm] d\left(\frac{y^2}{2x^2}+e^{xy}+Const\right)=0\longrightarrow\\[0.3cm] \frac{y^2}{2x^2}+e^{xy}+Const=Const_1\longrightarrow\boxed{\frac{y^2}{2x^2}+e^{xy}=c}$$

ANSWER

$$𝑦\left(𝑥^3\cdot 𝑒^{𝑥𝑦} − 𝑦\right)𝑑𝑥+𝑥\left(𝑦+𝑥^3\cdot 𝑒^{𝑥𝑦}\right)𝑑𝑦=0\longrightarrow\\[0.3cm] \frac{y^2}{2x^2}+e^{xy}=c$$

QUESTION 5

$$y^2dx+\left(x^2-xy-y^2\right)dy=0\longrightarrow\\[0.3cm] y^2dx=-x^2\cdot\left(1-\frac{y}{x}-\frac{y^2}{x^2}\right)dy\longrightarrow\\[0.3cm] \frac{dy}{dx}=\frac{-\displaystyle\frac{y^2}{x^2}}{1-\displaystyle\frac{y}{x}-\displaystyle\frac{y^2}{x^2}}\longrightarrow \left\{\begin{array}{l} y=xv\\[0.3cm] \displaystyle\frac{dy}{dx}=v+x\cdot\displaystyle\frac{dv}{dx} \end{array}\right.\\[0.3cm] v+x\cdot\frac{dv}{dx}=\frac{-v^2}{1-v-v^2}\longrightarrow\\[0.3cm] x\cdot\frac{dv}{dx}=\frac{-v\left(1-v-v^2\right)-v^2}{1-v-v^2}=\frac{-v+v^2+v^3-v^2}{1-v-v^2}\\[0.3cm] \left.x\cdot\frac{dv}{dx}=\frac{v^3-v}{1-v-v^2}\right|\cdot\left(\frac{\left(1-v-v^2\right)dx}{x\left(v^3-v\right)}\right)\\[0.3cm] \int\frac{\left(1-v^2\right)-v}{v\left(v^2-1\right)}dv=\int\frac{dx}{x}\longrightarrow\\[0.3cm] \int\left(-\frac{1}{v}-\frac{1}{2}\cdot\frac{(v+1)-(v-1)}{(v-1)(v+1)}\right)dv=\ln|x|+\ln|C|\\[0.3cm] \int\left(-\frac{1}{v}-\frac{1}{2}\cdot\frac{1}{(v-1)}+\frac{1}{2}\cdot\frac{1}{(v+1)}\right)dv=\ln|x|+\ln|C|\\[0.3cm] -\ln|v|-\frac{1}{2}\ln|v-1|+\frac{1}{2}\cdot\ln|v+1|=\ln|Cx|\\[0.3cm] \ln\left|\frac{1}{v}\cdot\sqrt{\frac{v+1}{v-1}}\right|=\ln|Cx|=\left[v=\frac{y}{x}\right]\\[0.3cm] \frac{x}{y}\cdot\sqrt{\frac{\displaystyle\frac{y}{x}+1}{\displaystyle\frac{y}{x}-1}}=Cx\longrightarrow\boxed{\frac{x}{y}\cdot\sqrt{\frac{y+x}{y-x}}=Cx}$$

ANSWER

$$y^2dx+\left(x^2-xy-y^2\right)dy=0\longrightarrow\\[0.3cm] \frac{x}{y}\cdot\sqrt{\frac{y+x}{y-x}}=Cx$$

QUESTION 7

$$x^4\cdot y'+x^3y=-\sec(xy)\longrightarrow\\[0.3cm] \text{Let's make a replacement}\,\,\,u=xy\longrightarrow y=\frac{u}{x}\longrightarrow\boxed{ y'=\frac{x\cdot u'-u}{x^2}}\\[0.3cm]$$

We substitute the replacement and the found derivative into the initial equation

$$x^4\cdot\frac{x\cdot u'-u}{x^2}+x^3\cdot\frac{u}{x}=-\frac{1}{\cos u}\longrightarrow\\[0.3cm] x^2\cdot\left(x\cdot u'-u+u\right)=-\frac{1}{\cos u}\longrightarrow\\[0.3cm] \left.x^3\cdot\frac{du}{dx}=-\frac{1}{\cos u}\right|\cdot\left(\frac{\cos udx}{x^3}\right)\\[0.3cm] \int\cos udu=\int\left(-\frac{dx}{x^3}\right)\longrightarrow\sin u=\frac{1}{2x^2}+c\longrightarrow\\[0.3cm] u\equiv xy=\arcsin\left(c+\frac{1}{2x^2}\right)\longrightarrow\\[0.3cm] \boxed{y=\frac{1}{x}\cdot\arcsin\left(c+\frac{1}{2x^2}\right)}$$

ANSWER

$$x^4\cdot y'+x^3y=-\sec(xy)\longrightarrow\\[0.3cm] y=\frac{1}{x}\cdot\arcsin\left(c+\frac{1}{2x^2}\right)$$

QUESTION 8

$$\left(1+\sin 𝑦\right)\cdot\frac{dx}{dy}=2𝑦\cos 𝑦−𝑥\left(\sec 𝑦+\tan 𝑦\right)\longrightarrow\\[0.3cm] y_{hom.}(x) :\quad\left(1+\sin 𝑦\right)\cdot\frac{dx}{dy}=-x\left(\frac{1}{\cos y}+\frac{\sin y}{\cos y}\right)\\[0.3cm] \left.\left(1+\sin 𝑦\right)\cdot\frac{dx}{dy}=-x\left(\frac{1+\sin y}{\cos y}\right)\right|\cdot\left(\frac{dy}{x\left(1+\sin y\right)}\right)\\[0.3cm] \int\frac{dx}{x}=-\int\frac{dy}{\cos y}\longrightarrow\ln|x|=-\int\frac{\cos y dy}{\cos^2y}\boxed{\longrightarrow}\\[0.3cm] \int\frac{\cos ydy}{\cos^2 y}=\int\frac{d\left(\sin y\right)}{1-\sin^2 y}=\\[0.3cm] =\int\frac{1}{2}\cdot\left(\frac{(1+\sin y)+(1-\sin y)}{(1-\sin y)(1+\sin y)}\right)d\left(\sin y\right)=\\[0.3cm] =\frac{1}{2}\cdot\int\left(\frac{1}{1-\sin y}+\frac{1}{1+\sin y}\right)d\left(\sin y\right)=\\[0.3cm] =\frac{1}{2}\cdot\left(-\ln|1-\sin y|+\ln|1+\sin y|\right)=\\[0.3cm] =\frac{1}{2}\cdot\ln\left|\frac{1+\sin y}{1-\sin y}\right|=\ln\left|\sqrt{\frac{\left(1+\sin y\right)^2}{(1-\sin y)(1+\sin y)}}\right|=\\[0.3cm] =\ln\left|\frac{1+\sin y}{\sqrt{1-\sin^2y}}\right|=\ln\left|\frac{1+\sin y}{\cos y}\right|\\[0.3cm] \boxed{\longrightarrow}\ln|x|=\ln|C|-\ln\left|\frac{1+\sin y}{\cos y}\right|\longrightarrow\\[0.3cm] \ln|x|=\ln\left|\frac{C\cos y}{1+\sin y}\right|\longrightarrow\boxed{x(y)=\frac{C\cos y}{1+\sin y}}$$

Let's apply the constant variation method $C\longrightarrow C(y) :$

$$\frac{dx}{dy}=\frac{\left(C'\cos y-C\sin y\right)\left(1+\sin y\right)-C\cos^2y}{\left(1+\sin y\right)^2}\\[0.3cm] \left(1+\sin y\right)\cdot\frac{dx}{dy}=2y\cos y-x\left(\frac{1+\sin y}{\cos y}\right)\\[0.3cm] \left(1+\sin y\right)\cdot\frac{\left(C'\cos y-C\sin y\right)\left(1+\sin y\right)-C\cos^2y}{\left(1+\sin y\right)^2}=\\[0.3cm] =2y\cos y-\frac{C\cos y}{1+\sin y}\cdot\frac{1+\sin y}{\cos y}\\[0.3cm] \frac{C'\cos y\left(1+\sin y\right)-C\sin y-C\sin^2y-C\cos^2}{1+\sin y}=2y\cos y-C\\[0.3cm] C'\cos y-C\cdot\frac{\sin y+\overbrace{\left(\sin^2 y+\cos^2 y\right)}^{=1}}{1+\sin y}=2y\cos y-C\\[0.3cm] C'\cos y-\cancel{C}\cdot\frac{\cancel{1+\sin y}}{\cancel{1+\sin y}}=2y\cos y-\cancel{C}\\[0.3cm] \left.\frac{dC}{dy}\cdot\cancel{\cos y}=2y\cdot\cancel{\cos y}\right|\cdot dy\\[0.3cm] \int dC=\int2ydy\longrightarrow\boxed{C(y)=y^2+c}$$

Conclusion,

$$x(y)=\frac{C(y)\cdot\cos y}{1+\sin y}\equiv\frac{\left(y^2+c\right)\cdot\cos y}{1+\sin y}$$

ANSWER

$$\left(1+\sin 𝑦\right)\cdot\frac{dx}{dy}=2𝑦\cos 𝑦−𝑥\left(\sec 𝑦+\tan 𝑦\right)\longrightarrow\\[0.3cm] x(y)=\frac{\left(y^2+c\right)\cdot\cos y}{1+\sin y}$$

QUESTION ( first line equation )

I just repeat the solution to equation # 2.

$$\left.\left(1-2e^{x/y}\right)dx-2e^{x/y}\cdot\left(1-\frac{x}{y}\right)dy=0\right|\div\left(dy\right)\\[0.3cm] \left(1-2e^{x/y}\right)\cdot\frac{dx}{dy}-2e^{x/y}\cdot\left(1-\frac{x}{y}\right)=0\\[0.3cm] \text{Let}\quad x=yv\longrightarrow\frac{dx}{dy}=v+y\cdot\frac{dv}{dy}\\[0.3cm] \left.\left(1-2e^v\right)\cdot\left(v+y\cdot\frac{dv}{dy}\right)-2e^v\cdot\left(1-v\right)=0\right|\div\left(1-2e^v\right)\\[0.3cm] v+y\cdot\frac{dv}{dy}=\frac{2e^v-2ve^v}{1-2e^v}\longrightarrow y\cdot\frac{dv}{dy}=\frac{2e^v-2ve^v}{1-2e^v}-v\\[0.3cm] y\cdot\frac{dv}{dy}=\frac{2e^v-2ve^v-v+2ve^v}{1-2e^v}\equiv\frac{2e^v-v}{1-2e^v}\\[0.3cm] \left.y\cdot\frac{dv}{dy}=\frac{2e^v-v}{1-2e^v}\right|\cdot\left(\frac{dy}{y}\cdot\frac{1-2e^v}{2e^v-v}\right)\\[0.3cm]\\[0.3cm] \int\frac{1-2e^v}{2e^v-v}dv=\int\frac{dy}{y}\longrightarrow\int\frac{d\left(v-2e^v\right)}{2e^v-v}=\int\frac{dy}{y}\\[0.3cm] -\ln\left|v-2e^v\right|=\ln|y|+\ln|C|\longrightarrow\\[0.3cm] \left.-\ln\left|y\cdot\left(v-2e^v\right)\right|=\ln|C|\right|\cdot\left(-1\right)\\[0.3cm] \ln\left|y\cdot\left(\frac{x}{y}-2e^{x/y}\right)\right|=\ln\left|\frac{1}{C}\right|\equiv\ln|c|\\[0.3cm] \boxed{x-2ye^{x/y}=c}$$

ANSWER

$$\left(1-2e^{x/y}\right)dx-2e^{x/y}\cdot\left(1-\frac{x}{y}\right)dy=0\\[0.3cm] x-2ye^{x/y}=c$$