التمارين العامة الخاصة بالفصل الخامس

(1)- حل المعادلة التفاضلية الآتية: $y^{\mathrm{\prime }}=\frac{{\cos }^{2}y}{x} , y=\frac{\pi }{4} , x=1$

$\begin{array}{rl}& \frac{dy}{dx}=\frac{{\cos }^{2}y}{x}\Rightarrow xdy={\cos }^{2}ydx\stackrel{(÷x{\cos }^{2}y)}{⟹}=\frac{dy}{{\cos }^{2}y}=\frac{dx}{x}\Rightarrow \int \frac{dy}{{\cos }^{2}y}=\int \frac{dx}{x}\\ & \int {\sec }^{2}ydy=\int \frac{dx}{x}\left(\frac{1}{{\cos }^{2}y}={\sec }^{2}y\right)\\ & \tan y=\ln \left|x\right|+c\left(\tan \frac{\pi }{4}=1,\ln 1=0\right)\\ & \tan \frac{\pi }{4}=\ln \left|1\right|+c\Rightarrow 1=0+c\Rightarrow c=1\\ & \therefore \tan y=\ln \left|x\right|+1\end{array}$

(2)- حل المعادلة التفاضلية $\frac{dy}{dx}=-2x\tan y$ حيث $x=0$ عندما $y=\frac{\pi }{2}$

$\begin{array}{rl}& \frac{dy}{dx}=-2x\tan y\Rightarrow [dy=-2x\tan ydx]÷\tan y\\ & \frac{dy}{\tan y}=-2xdx\Rightarrow \int \frac{dy}{\frac{\sin y}{\cos y}}=\int -2xdx\Rightarrow \int \frac{\cos y}{\sin y}dy=\int -2xdx\\ & \ln |\sin y|=-2\frac{x^2}{2}+c\Rightarrow \ln |\sin y|=-x^2+c\\ & \left(y=\frac{\pi }{2},x=0\right)\\ & \ln |\sin \frac{\pi }{2}|=0+c\Rightarrow \ln \left(1\right)=c\Rightarrow c=0\\ & \left(\sin \frac{\pi }{2}=1\right)\\ & \therefore \ln |\sin y|=-x^2\stackrel{\left(\mathrm{للطرفين} e \mathrm{بأخذ}\right)}{⟹}\sin y=e^{-x^2}\end{array}$

(3)- حل المعادلة التفاضلية $y\text{'}=2e^x{y}^3$ حيث أن $x=0$ عندما $y=\frac{1}{2}$

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

(4)- (حسب المنهج الجديد) حل المعادلة التفاضلية $xy\text{'}=y-x$ حيث أن $y=1,x=1$

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

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

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

نعوض $y=1,x=1$ لإيجاد قيمة الثابت $c$

$\begin{array}{rl}& \frac{y}{x}=-\ln \left|x\right|+c\Rightarrow \frac{1}{1}=-\ln \left|1\right|+c\Rightarrow 1=0+c\Rightarrow c=1\\ & \therefore \frac{y}{x}=-\ln \left|x\right|+1\end{array}$

(5)- حل المعادلة التفاضلية الآتية: $(x^2+3y^2)dx-2xydy=0$

نقسم البسط والمقام على $x^2\ne 0$

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

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

$\begin{array}{rl}& v+x\frac{dv}{dx}=\frac{1+3v^2}{2v}\Rightarrow x\frac{dv}{dx}=\frac{1+3v^2}{2v}-v\Rightarrow x\frac{dv}{dx}=\frac{1+3v^2-2v^2}{2v}\\ & x\frac{dv}{dx}=\frac{v^2+1}{2v}\Rightarrow xdv=\frac{v^2+1}{2v}dx\Rightarrow \frac{dv}{\frac{v^2+1}{2v}}=\frac{dx}{x}\Rightarrow \frac{2vdv}{v^2+1}=\frac{dx}{x}\\ & \int \frac{2vdv}{v^2+1}=\int \frac{dx}{x}\Rightarrow \ln |v^2+1|+\ln \left|c\right|=\ln \left|x\right|\Rightarrow \ln \left|x\right|=\ln \left|c(v^2+1)\right|\\ & x=c\cdot (v^2+1)\Rightarrow x=c(\frac{y^2}{x^2}+1)\Rightarrow x=c\left(\frac{y^2+x^2}{x^2}\right)\end{array}$