Section 9.3: Separable Differential Equations Lecture Video ↗
1. Find the general solution of the following differential equations:
-
(a)
\(\displaystyle\frac{dy}{dx}=6x\sqrt{y}\) Solution ↗
-
(b)
\(\displaystyle \frac{dy}{dx}= \frac{1+\sin(x)}{y}\) Solution ↗
-
(c)
\(\displaystyle \frac{dy}{dx}= \frac{x+1}{xy}\) Solution ↗
-
(d)
\(\displaystyle \frac{d u}{d t}=\frac{2+t^{4}}{u^3 t^{2}+u^{4} t^{2}}\) Solution ↗
2. Find the solution of the differential equation that satisfies the given initial condition:
-
(a)
\(\displaystyle \frac{dy}{dx}=2y, \quad y(1)= e\) Solution ↗
-
(b)
\(\displaystyle \frac{dy}{dx}= xe^y, \quad y(0)= 0\) Solution ↗
-
(c)
\(\displaystyle \frac{dy}{dx}=\frac{3}{1+x^2}, \quad y(1)=\pi\) Solution ↗
3. Applied Geometry Problem
Find an equation of the curve that passes through the point \((1, 3)\) and whose slope at \((x, y)\) is \(\dfrac{x}{y}\). Solution ↗
4. Cooling Model (Newton's Law of Cooling)
The differential equation below models the temperature of a \(93^\circ\text{C}\) cup of coffee in a \(17^\circ\text{C}\) room, where it is known that the coffee cools at a rate of \(1^\circ\text{C}\) per minute when its temperature is \(67^\circ\text{C}\). Solve the differential equation to find an expression for the temperature of the coffee at time \(t\). (Let \(y\) be the temperature in \(^\circ\text{C}\), and let \(t\) be the time in minutes, with \(t=0\) corresponding to \(93^\circ\text{C}\).)
\(\displaystyle \frac{dy}{dt}=-\frac{1}{50}(y-17)\) Solution ↗