تمارين (2-5)

تمارين (2-5)

(1)- حل المعادلات التفاضلية الآتية بطريقة فصل المتغيرات:

$y^{\mathrm{\prime }}{\cos }^{3}x=\sin x$

$\begin{array}{rl}& \frac{dy}{dx}\cdot {\cos }^{3}x=\sin x\Rightarrow ({\cos }^{3}x)dy=(\sin x)dx\Rightarrow dy=\frac{\sin x}{{\cos }^{3}x}dx\\ & dy=\frac{\sin x}{\cos x}\frac{1}{{\cos }^{2}x}dx\Rightarrow dy=\tan x{\sec }^{2}xdx\Rightarrow \int dy=\int \tan x{\sec }^{2}xdx\\ & y=\frac{(\tan x{)}^2}{2}+c\end{array}$

$\frac{dy}{dx}+xy=3xx=1,y=2$

$\begin{array}{rl}& \frac{dy}{dx}=3x-xy\Rightarrow \frac{dy}{dx}=x(3-y)\Rightarrow \frac{dy}{3-y}=xdx\\ & -1\int \frac{(-1)dy}{3-y}=\int xdx\Rightarrow -\ln |3-y|=\frac{x^2}{2}+cx=1y=2c=-\frac{1}{2}\\ & -\ln |3-2|=\frac{1}{2}+c\Rightarrow -\ln 1=\frac{1}{2}+c\Rightarrow 0=\frac{1}{2}+c\Rightarrow c=\frac{1}{2}-\frac{x^2}{2}\Rightarrow y=3-e^{\frac{1}{2}(1-x^2)}\\ & -\ln |3-y|=\frac{x^2}{2}-\frac{1}{2}\stackrel{\times -1}{\Rightarrow }\ln |3-y|=\frac{1}{2}-\frac{x^2}{2}\Rightarrow 3-y=e^{\frac{1}{2}}\Rightarrow y\end{array}$

$\frac{dy}{dx}=(x+1)(y-1)$

$\begin{array}{rl}& \frac{dy}{(y-1)}=(x+1)dx\Rightarrow \int \frac{dy}{(y-1)}=\int (x+1)dx\Rightarrow \ln |y-1|=\frac{x^2}{2}+x+c\\ & (y-1)=e^{\frac{x^2}{2}+x+c}\Rightarrow y=e^{\frac{x^2}{2}+x+c}+1\end{array}$

$(y^2+4y-1)y^{\mathrm{\prime }}=x^2-2x+3$

$\begin{array}{c}y\frac{dy}{dx}=4{(1+y^2)}^{\frac{3}{2}}\Rightarrow \frac{ydy}{{(1+y^2)}^{\frac{3}{2}}}=4dx\Rightarrow \int \frac{ydy}{{(1+y^2)}^{\frac{3}{2}}}=4\int dx\\ \int y{(1+y^2)}^{\frac{-3}{2}}\left(dy\right)=\int 4dx\Rightarrow \frac{1}{2}\int {(1+y^2)}^{\frac{-3}{2}}2ydy=\int 4dx\\ \frac{1}{2}\left(\frac{{(1+y^2)}^{\frac{-1}{2}}}{-\frac{1}{2}}\right)=4x+c\Rightarrow -{(1+y^2)}^{\frac{-1}{2}}=4x+c\Rightarrow \frac{-1}{\sqrt{1+y^2}}=4x+c\end{array}$

$e^x\cdot dx-y^{3}dy=0$

$\begin{array}{rl}& e^x\cdot dx-y^{3}dy=0\Rightarrow e^x\cdot dx=y^{3}dy\Rightarrow \int e^{x}dx=\int y^{3}dy\Rightarrow e^x=\frac{y^4}{4}+c\\ & \frac{y^4}{4}=e^x-c\Rightarrow y^4=4e^x-4c\Rightarrow y=\pm \sqrt{4e^x-4c}\Rightarrow y=\pm \sqrt{4e^x+c_1}(c_1=-4c)\end{array}$

$y^{\mathrm{\prime }}=2e^x{y}^{3}x=0,y=\frac{1}{2}$

$\begin{array}{rl}& \frac{dy}{dx}=2e^x{y}^3\Rightarrow dy=2e^x{y}^{3}dx\stackrel{\stackrel{y^3}{⟹}}{\frac{dy}{y^3}}=2e^{x}dx\Rightarrow \int \frac{dy}{y3}=\int 2e^{x}dx\\ & \int y^{-3}dy=2\int e^{x}dx\Rightarrow \frac{y^{-2}}{-2}=2e^x+c\Rightarrow \frac{-1}{2y^2}=2e^x+c\stackrel{x(-1)}{⟹}\frac{1}{y^2}=-4e^x-2c\\ & \frac{1}{{\left(\frac{1}{2}\right)}^2}=-4e^0-2c\Rightarrow 4=-4-2c\Rightarrow -2c=8\Rightarrow c=-4\\ & \frac{1}{y^2}=-4e^x+8\Rightarrow y^2=\frac{1}{-4e^x+8}\Rightarrow y=\pm \sqrt{\frac{1}{-4e^x+8}}\end{array}$

(2)- جد الحل العام للمعادلات التفاضلية الآتية:

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

$$\begin{gathered} \begin{array}{rl}& xy\frac{dy}{dx}=1-y^2-y^2\Rightarrow xy\frac{dy}{dx}=1-2y^2\Rightarrow xydy=(1-2y^2)dx\\ & \frac{ydy}{(1-2y^2)}=\frac{dx}{x}\Rightarrow \int \frac{ydy}{(1-2y^2)}=\int \frac{dx}{x}\Rightarrow -\frac{1}{4}\int \frac{(-4y)}{(1-2y^2)}dy=\int \frac{dx}{x}\end{array} \\ \begin{array}{rl}& -\frac{1}{4}\ln |1-2y^2|=\ln \left|x\right|+\ln \left|c\right|\Rightarrow \ln {(1-2y^2)}^{-\frac{1}{4}}=\ln \left|cx\right| \left(\mathrm{للطرفين} \ln \mathrm{بأخذ}\right)\\ & {(1-2y^2)}^{-\frac{1}{4}}=cx\Rightarrow \frac{1}{{(1-2y^2)}^{\frac{1}{4}}}=cx\Rightarrow \frac{1}{\sqrt[4]{1-2y^2}}=cx\Rightarrow \sqrt[4]{1-2y^2}=\frac{1}{cx}\\ & 1-2y^2=\frac{1}{(cx{)}^4}\Rightarrow 2y^2=1-\frac{1}{(cx{)}^4}\Rightarrow y^2=\frac{1}{2}-\frac{1}{2x^4{c}^4}\Rightarrow y=\pm \sqrt{\frac{1}{2}-\frac{1}{c_1{x}^4}}(2c^4=c_1)\end{array} \end{gathered}$$

$\sin x\cos y\frac{dy}{dx}+\cos x\sin y=0$

$\begin{array}{rl}& \sin x\cos y\frac{dy}{dx}=-\cos x\sin y\Rightarrow \frac{\cos y}{\sin y}\frac{dy}{dx}=-\frac{\cos x}{\sin x}\Rightarrow \frac{\cos y}{\sin y}dy=-\frac{\cos x}{\sin x}dx\\ & \int \frac{\cos y}{\sin y}dy=\int -\frac{\cos x}{\sin x}dx\Rightarrow \ln |\sin y|=-\ln |\sin x|+c\\ & \ln |\sin y|=\ln \left|\right(\sin x{)}^{-1}|+\ln c_1\Rightarrow \ln |\sin y|=\ln \left|c_1\right(\sin x{)}^{-1}|\\ & \ln \sin y=\ln \left|\frac{c_1}{\sin x}\right|\Rightarrow \sin y=\pm \frac{c_1}{\sin x}\end{array}$

$x{\cos }^{2}ydx+\tan ydy=0$

$\begin{array}{rl}& \tan ydy=-x{\cos }^{2}ydx\stackrel{÷{\cos }^{2}y}{⟹}\frac{\tan y}{{\cos }^{2}y}dy=-xdx\Rightarrow \int \frac{\tan y}{{\cos }^{2}y}dy=-\int xdx\\ & \int \tan y\cdot {\sec }^{2}ydy=-\int xdx\Rightarrow \frac{(\mathrm{tany}{)}^2}{2}=-\frac{x^2}{2}+c\stackrel{\times 2}{\Rightarrow }\frac{1}{2}(\tan y{)}^2=-\frac{1}{2}{x}^2+c\end{array}$

$ta{n}^{2}ydy=si{n}^{3}xdx$

$$\begin{gathered} {\tan }^{2}ydy={\sin }^{3}xdx\Rightarrow \int ({\sec }^{2}y-1)dy=\int {\sin }^{2}x\sin xdx \\ \begin{array}{rl}& \int ({\sec }^{2}y-1)dy=\int (1-{\cos }^{2}x)\sin xdx\\ & \int ({\sec }^{2}y-1)dy=\int [\sin x-(\cos x{)}^2\sin x]dx\\ & \int ({\sec }^{2}y)dy-\int dy=\int \sin xdx+\int (\cos x{)}^2(-\sin x)dx\\ & \tan y-y=-\cos x+\frac{{\cos }^{3}x}{3}+c\end{array} \end{gathered}$$

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

$\begin{array}{rl}& \frac{dy}{dx}={\cos }^{2}x{\cos }^{2}y\Rightarrow \frac{dy}{{\cos }^{2}y}={\cos }^{2}xdx\Rightarrow \int \frac{dy}{{\cos }^{2}y}=\int {\cos }^{2}xdx\\ & \int {\sec }^{2}ydy=\int \frac{1}{2}(1+\cos 2x)dx\Rightarrow \tan y=\frac{1}{2}(x+\frac{1}{2}\sin 2x)+c\end{array}$

$\frac{dy}{dx}=\frac{\cos x}{3y^2+e^y}$

$$\begin{gathered} \frac{dy}{dx}=\frac{\cos x}{3y^2+e^y}\Rightarrow 3y^2+e^{y}dy=\cos xdx\Rightarrow \int (3y^2+e^y)dy=\int \cos xdx \\ \frac{3y^3}{3}+e^y=\sin x+c\Rightarrow y^3+e^y=\sin x+c \end{gathered}$$

$e^{x+2y}+y^{\mathrm{\prime }}=0$

$$\begin{gathered} e^{x+2y}+y^{\mathrm{\prime }}=0\Rightarrow e^x\cdot e^{2y}+\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-e^x\cdot e^{2y}\Rightarrow \frac{dy}{e^{2y}}=-e^{x}dx \\ \begin{array}{c}\int \frac{dy}{e^{2y}}=-\int e^{x}dx\Rightarrow -\frac{1}{2}\int e^{-2y}(-2)dy=-\int e^{x}dx\\ -\frac{1}{2}{e}^{-2y}=-e^x+c\Rightarrow \frac{-1}{2e^{2y}}=-e^x+c\end{array} \end{gathered}$$