Separable Differential Equations

Exercise 1.2

1. Find the general solution of the following differential equations.

(a) $(\sin{x})y^{\prime} + (\cos{x})y = 0$
(b) $y^{\prime} = e^{x+y}$
(c) $y^{\prime} = \frac{1 - y^{2}}{1 - x^{2}}$
(d) $y^{\prime} = (1 + 2x)(1 + y^{2})$
2. Solve the following initial problem.
(a) $(1 + e^{x})y^{\prime} = y, \ y(0) = 1$
(b) $y^{\prime} + y\sin{x} = 0, \ y(\pi) = 3$
(c) $\ xyy^{\prime} - y^{2} = 1, \ y(2) = 1$
(d) $\ y^{\prime} = \frac{x(y^2 - 1)}{(x - 1)y^{3}}, \ y(2) = 2$ 3.
We took out the object which is $70^{\circ}c$ to the outside whose temperature is $20^{\circ}c$. After 15 minutes, the object temperature was $50^{\circ}c$. Answer the following questions
(a) Find the object's temperature after $30$ minutes.
(b) Find the time it takes for the temperature of object to reach $32^{\circ}c$.

Answer
1. (a) Separate the variables.

$$\frac{y^{\prime}}{y} = - \frac{\cos{x}}{\sin{x}}$$

Then integrate both sides.

$$\int \frac{dy}{y} = - \int \frac{\cos{x}}{\sin{x}}dx$$

Then we have

$$\log{\vert y\vert} = - \log{\vert\sin{x}\vert} + c$$

Now we take the exponenetial.

$$\vert y\vert = e^{- \log{\vert\sin{x}\vert} + c} = e^{- \log{\vert\sin{x}\vert}} \cdot e^{c}$$

$c$ is an arbitrary constant. So, we write $C$ for $e^{c}$. Therefore,

$$\vert y\vert = C \frac{1}{\vert\sin{x}\vert}$$

Now taking out the absolute value sign, we have $y = \pm C \frac{1}{\vert\sin{x}\vert}$. But $\pm C$ is again a constant. So, we use $C$.

$$y = C \frac{1}{\vert\sin{x}\vert}$$

Lastly, taking out the absolute sign of $\sin{x}$, then we have $y = \pm C \sin{x}$. But $\pm C$ is again a constant. So, we use $C$. Then

$$y = C \frac{1}{\sin{x}}$$

Hence,

$$(\sin{x})y = C$$

Representing different values using the same $C$ in this way is called abuse of $C$. From now on we do use the abuse of $C$ without telling you.

(b) Separating the variables, we have

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

Now integrate both sides,

$$\int e^{-y} dy = \int e^{x}dx$$

Then

$$- e^{-y} = e^{x} + c$$

or

$$e^{-y} = - e^{x} + C$$

Taking the logarithm, we have

$$-y = \log{(C - e^{x})}$$

Therefore,

$$y = - \log{(C - e^{x})} \ \ \$$

(c) Separate the variables and integrate both sides, we have

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

Thus,

$$\frac{1}{2}\log{\vert\frac{1+y}{1-y}\vert} = \frac{1}{2}\log{\vert\frac{1+x}{1-x}\vert} + c$$

From this, we have

$$\log{\vert\frac{1+y}{1-y}\vert} = \log{\vert\frac{1+x}{1-x}\vert} + C$$

Now take logarithm and accept the abuse of $C$, we have

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

Write this for $y$,

$$(1 + y)(1 - x) = C(1 + x)(1 - y)$$

and

$$y[(1 - x) + C(1 + x) = C(1+x) - 1 + x$$

Therefore,

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

(d) Separate the variables and integrate both sides,

$$\int \frac{dy}{1 + y^{2}} = \int (1 + 2x) dx$$

Thus,

$$\arctan{y} = x + x^2 + c$$

$$y = \tan(x + x^2 + c) \ \ \$$

2. (a) Separate the variables and integrate both sides,

$$\int \frac{dy}{y} = \int \frac{dx}{1 + e^{x}} = \int \frac{e^{-x} dx}{1 + e^{-x}}$$

Then

$$\log{\vert y\vert} = - \log{\vert 1 + e^{-x}\vert} + c$$

Thus,

$$y = \frac{C}{1+e^{-x}}$$

Now applying the initial value $y(0) = 1$. Then $1 = \frac{C } {1 + 1}$ and

$$y = \frac{2}{1 + e^{-x}} \ \ \$$

(b) Separate the variables and integrate both sides,

$$\int \frac{dy}{y} = - \int \sin{x} dx$$

and

$$\log{\vert y\vert} = \cos{x} + c$$

Thus,

$$y = Ce^{\cos{x}}$$

Now using the initial condition $y(\pi) = 3$, we have $3 = C e^{-1}$ and

$$y = 3 e^{\cos{x} + 1} \ \ \$$

(c) Separate the variables and integrate both sides,

$$\int \frac{y dy}{1 + y^{2}} = \int \frac{1}{x} dx$$

and

$$\frac{1}{2}\log{\vert 1 + y^{2}\vert} = \log{\vert x\vert} + c$$

Thus,

$$\log{\vert 1 + y^{2}\vert} = 2 \log{\vert x\vert} + C$$

Now taking logarithm and accept the abuse of $C$, we have

$$1 + y^{2} = C x^{2}$$

Here using the initial condition $y(2) = 1$, we have $1 + 1 = C (2^{2})$ and

$$1 + y^{2} = \frac{1}{2}x^{2}$$

(d) Separating the variables, we have

$$\int \frac{y^{3}}{y^{2} -1} dy = \int \frac{x}{x - 1} dx$$

Integrate both sides,

$$\int (y + \frac{y}{y^{2} - 1}) dy = \int (1 + \frac{1}{x - 1})dx$$

and

$$\frac{y^2}{2} + \frac{1}{2}\log{\vert y^2 - 1\vert} = x + \log{\vert x - 1\vert} + c$$

Thus,

$$y^2 + \log{\vert y^2 - 1\vert} = 2x + \log{(x-1)^2} + C$$

Now use the initial condition $y(2) = 2$. Then $4 + \log{3} = 4 + \log{1} + C$implies that $C = \log{3}$. Therefore,

$$y^2 + \log{\vert y^2 - 1\vert} = 2x + \log{(x-1)^2} + \log{3} \ \ \$$

3. (a) Using the Newton's cooling law, we have

$$\frac{dT}{dt} = - \kappa(T - 20), \ T(0) = 70, \ T(15) = 50$$

Separate the variables, we have

$$\int \frac{dT}{T - 20} = \int - \kappa dt$$

and

$$\log{\vert T - 20\vert} = - \kappa t + c$$

Now taking the exponential, we can write

$$T - 20 = C e^{- \kappa t}$$

Next we use the boundary conditions $T(0) = 70, \ T(15) = 50$. Then $70 -20 = C, \ 50 - 20 = C e^{- 15 \kappa}$より $C = 50, \ e^{-\kappa} = (\frac{3}{5})^{1/15}$. Thus,

$$T = 20 + 50 (\frac{3}{5})^{\frac{t}{15}}$$

Therefore the temperature of the object after 30 minutes is given by

$$T(30) = 20 + 50 (\frac{3}{5})^{\frac{30}{15}} = 38^{\circ}C\ \ \$$

(b) Since the object's temperature is $32^{\circ}$,

$$32 = 20 + 50 (\frac{3}{5})^{\frac{t}{15}}$$

Now solve this for $t$. Then

$$\frac{12}{50} = (\frac{3}{5})^{\frac{t}{15}}$$

Taking the logarithm,

$$\log{\frac{12}{50} } = \frac{t}{15}\log{(\frac{3}{5})}$$

Therefore,

$$t = 15\frac{\log{(\frac{6}{25})}}{\log{(\frac{3}{5})}} = 41.9min \ \$$