Chapter 1: Problem 67
Solve $$ \sec ^{2} y \frac{d y}{d x}+\frac{1}{2 \sqrt{1+x}} \tan y=\frac{1}{2 \sqrt{1+x}} $$
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Chapter 1: Problem 67
Solve $$ \sec ^{2} y \frac{d y}{d x}+\frac{1}{2 \sqrt{1+x}} \tan y=\frac{1}{2 \sqrt{1+x}} $$
These are the key concepts you need to understand to accurately answer the question.
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Solve the given differential equation. $$y^{\prime}-x^{-1} y=4 x^{2} y^{-1} \cos x, \quad x>0$$
An object of mass \(m\) is released from rest in a medium in which the frictional forces are proportional to the square of the velocity. The initial- value problem that governs the subsequent motion is $$m v \frac{d v}{d y}=m g-k v^{2}, \quad v(0)=0$$ where \(v(t)\) denotes the velocity of the object at time \(t, y(t)\) denotes the distance travelled by the object at time \(t\) as measured from the point at which the object was released, and \(k\) is a positive constant.(a) Solve \((1.12 .6)\) and show that $$ v^{2}=\frac{m g}{k}\left(1-e^{-2 k y / m}\right) $$ (b) Make a sketch of \(v^{2}\) as a function of \(y\)
Consider the phenomenon of exponential decay. This occurs when a population \(P(t)\) is governed by the differential equation $$\frac{d P}{d t}=k P$$ where \(k\) is a negative constant. A population of swans in a wildlife sanctuary is declining due to the presence of dangerous chemicals in the water. If the population of swans is experiencing exponential decay, and if there were 400 swans in the park at the beginning of the summer and 340 swans 30 days later, (a) how many swans are in the park 60 days after the start of summer? 100 days after the start of summer? (b) how long does it take for the population of swans to be cut in half? (This is known as the half-life of the population.)
Solve the given differential equation. $$y y^{\prime}=\sqrt{x^{2}+y^{2}}-x, \quad x>0$$
Consider the differential equation $$ f^{\prime}(y) \frac{d y}{d x}+p(x) f(y)=q(x) $$ where \(p\) and \(q\) are continuous functions on some interval \((a, b),\) and \(f\) is an invertible function. Show that Equation \((1.8 .28)\) can be written as $$ \frac{d u}{d x}+p(x) u=q(x) $$ where \(u=f(y)\) and hence show that the general solution to Equation \((1.8 .28)\) is $$ y(x)=f^{-1}\left\\{I^{-1}\left[\int I(x) q(x) d x+c\right]\right\\} $$ where \(I\) is given in \((1.8 .27), f^{-1}\) is the inverse of \(f\) and \(c\) is an arbitrary constant.
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