Question

solve the following differential equation

1. dx+(1-x²)dy=0
2. 8cos²ydx+cosec²xdy=0

The annual sells of a new company are expected to grow at a rate proportional to the difference between the sells and the upper limit of 20 million pounds . If the sells are 0 initially and 4million pounds after the 2^nd year of operations. Determine if the companies sells during the 10^th year and the year when the company sells will be 15million pounds.

Answer 1

1.

$$dx+(1-x^2)dy=0$$

$$dy=\dfrac{dx}{x^2-1}$$

$$y=\int \dfrac{dx}{x^2-1}$$

$$y=\dfrac{1}{2}\int \dfrac{dx}{x-1}-\dfrac{1}{2}\int \dfrac{dx}{x+1}$$

$$y=\dfrac{1}{2}\ln |x-1|-\dfrac{1}{2}\ln|x+1|+C$$

2.

$$8\cos^2(y)dx+\cosec^2(x)dy=0$$

$$\dfrac{dy}{\cos^2(y)}=-8\sin^2(x)dx$$

$$\int\dfrac{dy}{\cos^2(y)}=-\int8\sin^2(x)dx$$

$$\int\dfrac{dy}{\cos^2(y)}=-4\int(1-\cos(2x))dx$$

$$\tan(y)=-4x+2\sin(2x)+C$$

3.

$$\dfrac{dy}{dt}=k(y-20)$$ $$\dfrac{dy}{y-20}=kdt$$

$$\int\dfrac{dy}{y-20}=\int kdt$$

$$\ln(y-20)=kt+\ln C$$

$$y=20+Ce^{kt}$$

$$y(0)=0:0=20+Ce^{k(0)}=>C=-20$$

$$y=20-20e^{kt}$$

$$y(2)=4:4=20-20e^{k(2)}=>e^{2k}=0.8$$

$$=>k=0.5\ln0.8$$

$$y(t)=20-20e^{(0.5\ln0.8)t}$$

$$y(10)=20-20e^{(0.5\ln0.8)\cdot10}$$

$$=13.4464\text{ million pounds }$$

$$y(t_1)=15:15=20-20e^{(0.5\ln0.8)t_1}$$

$$e^{(0.5\ln0.8)t_1}=0.25$$

$$(0.5\ln0.8)t_1=\ln0.25$$

$$t_1=2(\dfrac{\ln0.25}{\ln0.8})$$

$$t_1\approx12.425\ years$$