Practice for Solving by Direct Integration

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Why?

Solve $\frac {dy}{dx} = x^2+x$ with $y(1)=3$ . $y(x) = \answer {\frac {x^3}{3} + \frac {x^2}{2} + \frac {13}{6}}$

Solve $\frac {dy}{dx} = \sin (5x)$ with $y(0)=2$ . $y(x) = \answer {-\frac {1}{5}\cos (5x) + \frac {11}{5}}$

Solve $\frac {dy}{dx} = e^x + x$ with $y(0) = 10$ . $y(x) = \answer {e^x + \frac {x^2}{2} + 9}$

Solve $\frac {dy}{dx} = 2xe^{3x}$ with $y(0) = 1$ . $y(x) = \answer {\frac {2}{3}xe^{3x} - \frac {2}{9}e^{2x} + \frac {11}{9}}$

Solve $\frac {dx}{dt} = e^t\cos (2t) + t$ with $x(0) = 3$ . $x(t) = \answer {\frac {1}{5}e^t\cos (2t) + \frac {2}{5}e^t\sin (2t) + \frac {t^2}{2} + \frac {14}{5}}$

Solve $\frac {dy}{dx} = \frac {1}{x^2 + 1} + 3e^{2x}$ with $y(0) =2$ . $y(x) = \answer {\tan ^{-1}(x) + \frac {3}{2}e^{2x} + \frac {1}{2}}$

Solve $\frac {dy}{dx} = \frac {1}{x^2-1}$ for $y(0)=0$ . (This requires partial fractions or hyperbolic trigonometric functions.) $y(x) = \answer {\frac {1}{2}\ln \left (\frac {1-x}{1+x}\right )}$

[harder] Solve $y'' = \sin x$ for $y(0)=0$ , $y'(0) = 2$ . $y(x) = \answer {-\sin (x) + 3x}$

A spaceship is traveling at the speed $2t^2+1$ ^km/_s ( $t$ is time in seconds). It is pointing directly away from earth and at time $t=0$ it is 1000 kilometers from earth. How far from earth is it at one minute from time $t=0$ ? $x(60) = \answer {3460}$

Sid is in a car traveling at speed $10t+70$ miles per hour away from Las Vegas, where $t$ is in hours. At $t=0$ , Sid is 10 miles away from Vegas. How far from Vegas is Sid 2 hours later? $\answer {170}$

Solve $\frac {dx}{dt} = \sin (t^2)+t$ , $x(0)=20$ . It is OK to leave your answer as a definite integral. $x = \answer {\frac {t^2}{2} + \int _0^t \sin (s^2)ds + 20}$

Solve $\frac {dy}{dt} = e^{t^2} + \sin (t)$ , $y(0) = 4$ . The answer can be left as a definite integral. $y = \answer {-\cos (t) + \int _0^t e^{s^2} ds + 5}$

A dropped ball accelerates downwards at a constant rate $9.8$ meters per second squared. Set up the differential equation for the height above ground $h$ in meters. Then supposing $h(0) = 100$ meters, how long does it take for the ball to hit the ground. (Enter the exact value, not a decimal approximation) $t = \answer {\sqrt {\frac {100}{4.9}}}$

The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. The volume (in centimeters cubed) of a ball of radius $r$ centimeters is $(\frac {4}{3}) \pi r^3$ . The surface area is $4 \pi r^2$ . Set up the differential equation for how the radius $r$ is changing. (Use $C$ for an unknown constant) $r' = \answer {-C}$

Now, suppose that at time $t=0$ minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time $t$ will the snowball be completely melted? $\answer {25}$ minutes from $t=0$ .

How many distinct constants do you need for the general solution to $y''''= 0$ ? $\answer {4}$