حل كلاً من المعادلات التفاضلية الآتية:

$x(\frac{dy}{dx}-\tan \frac{y}{x})=y$

$\begin{array}{rl}& (\frac{dy}{dx}-\tan \frac{y}{x})=\frac{y}{x}\Rightarrow \frac{dy}{dx}=\frac{y}{x}+\tan \frac{y}{x}\dots \dots \dots \left(1\right)\\ & \frac{dy}{dx}=v+\tan v\dots \dots \dots \left(2\right)(v=\frac{y}{x} \left(\mathrm{وضعنا}\right)\text{)}\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \left(3\right)\end{array}$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$\begin{array}{rl}& v+x\frac{dv}{dx}=v+\tan v\Rightarrow x\frac{dv}{dx}=\tan v\Rightarrow \frac{dv}{\tan v}=\frac{dx}{x}\\ & \int \frac{dx}{x}=\int \frac{1}{\frac{\sin v}{\cos v}}dv\Rightarrow \int \frac{dx}{x}=\int \frac{\cos v}{\sin v}dv\Rightarrow \ln \left|x\right|=\ln |\sin v|+\ln \left|c\right|\\ & \ln \left|x\right|=\ln \left|c\right(\sin v\left)\right|\Rightarrow \left|x\right|=\left|c\right(\sin v\left)\right|\Rightarrow x=\pm c(\sin v)\Rightarrow x=\pm c(\sin \frac{y}{x})\end{array}$

تمارين (3-5)

تمارين (3-5)

(1)- حل كلاً من المعادلات التفاضلية الآتية:

$y^{\mathrm{\prime }}=\frac{y}{x}+e^{\frac{y}{x}}$

$\begin{array}{c}\frac{dy}{dx}=\frac{y}{x}+e^{\frac{y}{x}}[v=\frac{y}{x} \left(\mathrm{وضعنا}\right)]\\ \frac{dy}{dx}=v+e^v\dots \dots \dots \left(1\right)\\ y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \left(2\right)\end{array}$

نعوض المعادلة (2) في المعادلة (1) فينتج:

$$\begin{gathered} v+x\frac{dv}{dx}=v+e^v\Rightarrow x\frac{dv}{dx}=e^v\Rightarrow xdv=e^{v}dx\Rightarrow \frac{dv}{e^v}=\frac{dx}{x} \\ \int e^{-v}dv=\int \frac{dx}{x}\Rightarrow -e^{-v}+c=\ln \left|x\right|\Rightarrow \ln \left|x\right|=-e^{-\frac{y}{x}}+c \end{gathered}$$

$(y^2-xy)dx+x^{2}dy=0$

بقسمة البسط والمقام على $x^2\ne 0$

$$\begin{gathered} x^{2}dy=-(y^2-xy)dx\Rightarrow x^{2}dy=(xy-y^2)dx \\ \frac{dy}{dx}=\frac{xy-y^2}{x^2} \\ \begin{array}{rl}& \frac{dy}{dx}=\frac{\frac{xy}{x^2}-\frac{y^2}{x^2}}{\frac{x^2}{x^2}}\Rightarrow \frac{dy}{dx}=\frac{\frac{y}{x}-{\left(\frac{y}{x}\right)}^2}{1}\Rightarrow \frac{dy}{dx}=\frac{y}{x}-{\left(\frac{y}{x}\right)}^2\dots \dots .\left(1\right)\\ & \frac{dy}{dx}=v-(v{)}^2\dots \dots \dots (2)(v=\frac{y}{x} \left(\mathrm{وضعنا}\right))\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array} \end{gathered}$$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$\begin{array}{rl}& \therefore v+x\frac{dv}{dx}=v-v^2\Rightarrow x\frac{dv}{dx}=-v^2\Rightarrow xdv=-v^{2}dx\Rightarrow \frac{dv}{v^2}=-\frac{dx}{x}\\ & \int \frac{dv}{v^2}=\int \frac{dx}{x}\Rightarrow \int v^{-2}dv=-\int \frac{dx}{x}\Rightarrow \frac{v^{-1}}{-1}+c=-\ln \left|x\right|\\ & -\ln \left|x\right|=\frac{-1}{v}+c\stackrel{(x-1)}{⟹}\ln \left|x\right|=\frac{1}{\frac{y}{x}}-c\Rightarrow \ln \left|x\right|=\frac{x}{y}-c\Rightarrow \ln \left|x\right|=\frac{x}{y}+c_1(-c=c_1)\end{array}$

$(x+2y)dx+(2x+3y)dy=0$

نقسم البسط والمقام على $x\ne 0$

$\begin{array}{rl}& (2x+3y)dy=-(x+2y)dx\Rightarrow \frac{dy}{dx}=\frac{-x-2y}{2x+3y}\\ & \frac{dy}{dx}=\frac{\frac{-x}{x}-\frac{2y}{x}}{\frac{2x}{x}+3\frac{y}{x}}\Rightarrow \frac{dy}{dx}=\frac{-1-2\frac{y}{x}}{2+3\frac{y}{x}}\dots \cdots \cdots \left(1\right)\\ & \frac{dy}{dx}=\frac{-1-2v}{2+3v}\dots \dots \dots \left(2\right)(v=\frac{y}{x} \left(\mathrm{وضعنا}\right))\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \text{(3)}\end{array}$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$$\begin{gathered} v+x\frac{dv}{dx}=\frac{-1-2v}{2+3v}\Rightarrow x\frac{dv}{dx}=\frac{-1-2v}{2+3v}-v\Rightarrow x\frac{dv}{dx}=\frac{-1-2v-v(2+3v)}{2+3v} \\ \begin{array}{rl}& x\frac{dv}{dx}=\frac{-1-2v-2v-3v^2}{2+3v}\Rightarrow x\frac{dv}{dx}=\frac{-(3v^2+4v+1)}{2+3v}\\ & -xdv=\frac{3v^2+4v+1}{2+3v}dx\Rightarrow \int \frac{(2+3v)dv}{3v^2+4v+1}=-\int \frac{dx}{x}\Rightarrow \frac{1}{2}\int \frac{2(2+3v)dv}{3v^2+4v+1}=-\int \frac{dx}{x}\\ & \frac{1}{2}\ln |3v^2+4v+1|=-\ln \left|x\right|+c\Rightarrow -c=\ln \left|{(3v^2+4v+1)}^{\frac{1}{2}}\right|+\ln \left|x\right|\\ & {\ln }_{c_1}=\ln \left|x\sqrt{3v^2+4v+1}\right|\Rightarrow c_1=\left|x\sqrt{3v^2+4v+1}\right|\\ & c_1=\left|x\sqrt{\frac{y^2}{x^2}+4\frac{y}{x}+1}\right|\Rightarrow c_1=\left|x\sqrt{\frac{y^2+4xy+3y^2}{x^2}\mid }\right|\\ & c_1=\left|x\frac{\sqrt{y^2+4xy+3y^2}}{\sqrt{x^2}}\right|\Rightarrow c_1=\left|x\right|\left|\frac{\sqrt{y^2+4xy+3y^2}}{\left|x\right|}\right|\\ & c_1=\left|\sqrt{y^2+4xy+3y^2}\right|\end{array} \end{gathered}$$

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

نقسم البسط والمقام على $x^2\ne 0$

$\begin{array}{rl}& \begin{array}{rl}& \frac{dy}{dx}=\frac{\frac{x^2}{x^2}+\frac{y^2}{x^2}}{\frac{2xy}{x^2}}=\frac{1+{\left(\frac{y}{x}\right)}^2}{2\frac{y}{x}}\dots \dots \dots \left(1\right)\\ & \frac{dy}{dx}=\frac{1+v^2}{2v}\dots \dots \dots \left(2\right)(v=\frac{y}{x} \left(\mathrm{وضعنا}\right))\end{array}\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \left(3\right)\end{array}$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$$\begin{gathered} \begin{array}{rl}& v+x\frac{dv}{dx}=\frac{1+v^2}{2v}\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{2v}-v\Rightarrow x\frac{dv}{dx}=\frac{1+v^2-2v^2}{2v}\\ & x\frac{dv}{dx}=\frac{1-v^2}{2v}\Rightarrow xdv=\frac{1-v^2}{2v}dx\Rightarrow \frac{2vdv}{1-v^2}=\frac{dx}{x}\Rightarrow -\int \frac{-2vdv}{(1-v^2)}=\int \frac{dx}{x}\\ & -\ln \left|(1-v^2)\right|+\ln \left|c\right|=\ln \left|x\right|\Rightarrow \ln \left|(1-v^2)\right|+\ln \left|x\right|=\ln \left|c\right|\\ & \ln \left|x(1-v^2)\right|=\ln \left|c\right|\Rightarrow \left|c\right|=\left|x(1-v^2)\right|\end{array} \\ c=\pm x(1-v^2)\Rightarrow c=\pm x(1-\frac{y^2}{x^2})\Rightarrow c=\pm x\frac{x^2-y^2}{x^2}\Rightarrow c=\pm \left(\frac{x^2-y^2}{x}\right) \end{gathered}$$

$(y^2-x^2)dx+xydy=0\Rightarrow xydy=-(y^2-x^2)dx$

نقسم البسط والمقام على $x^2\ne 0$

$\begin{array}{rl}& \frac{dy}{dx}=\frac{x^2-y^2}{xy}\\ & \frac{dy}{dx}=\frac{\frac{x^2}{x^2}-\frac{y^2}{x^2}}{\frac{xy}{x^2}}\Rightarrow \frac{dy}{dx}=\frac{1-{\left(\frac{y}{y}\right)}^2}{\frac{y}{x}}\dots \dots \dots \left(1\right)\\ & \frac{dy}{dx}=\frac{1-v^2}{v}\dots \dots \left(2\right)(v=\frac{y}{1} \left(\mathrm{وضعنا}\right))\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \left(3\right)\end{array}$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$\begin{array}{rl}& v+x\frac{dv}{dx}=\frac{1-v^2}{v}\Rightarrow x\frac{dv}{dx}=\frac{1-v^2}{v}-v\Rightarrow x\frac{dv}{dx}=\frac{1-v^2-v^2}{v}\Rightarrow x\frac{dv}{dx}=\frac{1-2v^2}{v}\\ & \frac{vdv}{1-2v^2}=\frac{dx}{x}\Rightarrow \frac{-1}{4}\int \frac{(-4)vdv}{1-2v^2}=\int \frac{dx}{x}\Rightarrow \frac{-1}{4}\ln |1-2v^2|+\ln \left|c\right|=\ln \left|x\right|\\ & -\ln {|1-2v^2|}^{\frac{1}{4}}+\ln \left|c\right|=\ln \left|x\right|\Rightarrow \ln \left|{(1-2v^2)}^{\frac{1}{4}}\right|+\ln \left|x\right|=\ln \left|c\right|\\ & \ln \left|x{(1-2v^2)}^{\frac{1}{4}}\right|=\ln \left|c\right|\Rightarrow \ln \left|c\right|=\left|x^4\sqrt{1-2v^2}\right|\Rightarrow c=\pm x^4\sqrt{1-2v^2}\\ & \therefore c=\pm x\sqrt[4]{1-2{\left(\frac{y}{x}\right)}^2}\Rightarrow c=\pm x\sqrt[4]{1-\frac{2y^2}{x^2}}\Rightarrow c=\pm x\sqrt[4]{\frac{x^2-2y^2}{x^2}}\end{array}$

$x^{2}ydx=(x^3+y^3)dy$

نقسم البسط والمقام على $x^3\ne 0$

$\begin{array}{rl}& \frac{dy}{dx}=\frac{x^{2}y}{x^3+y^3}\\ & \frac{dy}{dx}=\frac{\frac{x^{2}y}{x^3}}{\frac{x^3+y^3}{x^3}}\Rightarrow \frac{dy}{dx}=\frac{\frac{x^{2}y}{x^3}}{\frac{x^3}{x^3}+\frac{y^3}{x^3}}\Rightarrow \frac{dy}{dx}=\frac{\frac{y}{x}}{1+{\left(\frac{y}{x}\right)}^3}\dots \dots \dots \left(1\right)\\ & \frac{dy}{dx}=\frac{v}{1+v^3}\dots \dots \dots \left(2\right)(v=\frac{y}{x} \left(\mathrm{وضعنا}\right))\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \left(3\right)\end{array}$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$\begin{array}{rl}& v+x\frac{dv}{dx}=\frac{v}{1+v^3}\Rightarrow x\frac{dv}{dx}=\frac{v}{1+v^3}-v\Rightarrow x\frac{dv}{dx}=\frac{v-v(1+v^3)}{1+v^3}\\ & x\frac{dv}{dx}=\frac{v-v-v^4}{1+v^3}\Rightarrow x\frac{dv}{dx}=\frac{-v^4}{1+v^3}\Rightarrow \frac{1+v^3}{v^4}dv=\frac{-dx}{x}\\ & \int \frac{1+v^3}{v^4}dv=-\int \frac{dx}{x}\Rightarrow \int \frac{1}{v^4}dv+\int \frac{v^3}{v^4}dv=-\int \frac{dx}{x}\\ & -\int \frac{dx}{x}=\int v^{-4}dv+\int \frac{1}{v}dv\Rightarrow -\ln \left|x\right|=\frac{1}{-3}{v}^{-3}+\ln \left|v\right|+\ln \left|c\right|\\ & -\ln \left|x\right|=\frac{-1}{3v^3}+\ln \left|v\right|+\ln \left|c\right|\Rightarrow \frac{1}{3v^3}=\ln \left|x\right|+\ln \left|v\right|+\ln \left|c\right|\\ & \frac{1}{3v^3}=\ln \left|xvc\right|+c\Rightarrow \frac{1}{3{\left(\frac{y}{x}\right)}^3}=\ln \left|x\frac{y}{x}c\right|\Rightarrow \frac{1}{3\frac{y^3}{x^3}}=\ln \left|yc\right|\\ & \frac{x^3}{3y^3}=\ln \left|yc\right|\Rightarrow y^3=\frac{x^3}{3\ln \left|cy\right|}\Rightarrow y=\frac{x}{\sqrt[3]{3\ln \left|cy\right|}}\end{array}$

$x(\frac{dy}{dx}-\tan \frac{y}{x})=y$

$\begin{array}{rl}& (\frac{dy}{dx}-\tan \frac{y}{x})=\frac{y}{x}\Rightarrow \frac{dy}{dx}=\frac{y}{x}+\tan \frac{y}{x}\dots \dots \dots \left(1\right)\\ & \frac{dy}{dx}=v+\tan v\dots \dots \dots \left(2\right)(v=\frac{y}{x} \left(\mathrm{وضعنا}\right)\text{)}\\ & \because y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\dots \dots \dots \left(3\right)\end{array}$

نعوض المعادلة (3) في المعادلة (2) فينتج:

$\begin{array}{rl}& v+x\frac{dv}{dx}=v+\tan v\Rightarrow x\frac{dv}{dx}=\tan v\Rightarrow \frac{dv}{\tan v}=\frac{dx}{x}\\ & \int \frac{dx}{x}=\int \frac{1}{\frac{\sin v}{\cos v}}dv\Rightarrow \int \frac{dx}{x}=\int \frac{\cos v}{\sin v}dv\Rightarrow \ln \left|x\right|=\ln |\sin v|+\ln \left|c\right|\\ & \ln \left|x\right|=\ln \left|c\right(\sin v\left)\right|\Rightarrow \left|x\right|=\left|c\right(\sin v\left)\right|\Rightarrow x=\pm c(\sin v)\Rightarrow x=\pm c(\sin \frac{y}{x})\end{array}$