Answer to Question #181852 in Calculus for sdfafds

Question

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1) The quantity of a substance can be modeled by the function k(t) that satisfies the differential equation dk/ dt =-1/90 (k-450). One point on this function is k(2) = 900. Based on this model, use a linear approximation to the graph of k(t) at t = 2 to estimate the quantity of the substance at t = 2.1.

2) Consider the differential equation dy/dx =x^2y^2with a particular solution y=f(x) having an initial condition y(−3) = 1 . Use the equation of the line tangent to the graph of f at the point (−3, 1) in order to approximate the value of f(−2.8) .

3) Given the differential equation dy/dx=-x/y^2 find the particular solution, y=f(x) , with the initial condition f(−6) = −3 .

4) What is the particular solution to the differential equation dy/dx =3 cos(x)y with the initial condition y(pi/2)=-2?

Answer 1

Solution.

1)

$\frac{dk}{dt}=-\frac{1}{90}(k-450), k(2)=900, k(2.1)=?$

$\frac{dk}{dt}=-\frac{1}{90}(k--450)=5-\frac{1}{90}k$

$k'+\frac{1}{90}k=5,\newline k'+\frac{1}{90}k==0,\newline \frac{dk}{k}=-\frac{1}{90}, \newline k=Ce^{-\frac{1}{90}}t,$ where $C$ is some constant.

$k=C(t)e^{-\frac{1}{90}}t,\newline C'(t)e^{-\frac{1}{90}}t-\frac{1}{90}C(t)e^{-\frac{1}{90}}t+\frac{1}{90}C(t)e^{-\frac{1}{90}}t=5, \newline dC(t)=5e^{\frac{1}{90}t}dt.\newline C(t)=450e^{\frac{1}{90}}t$

So,

$k(t)=Ce^{-\frac{t}{90}}+450.\newline \text{If } k(2)=900, \text{then} \newline 900=Ce^{-\frac{1}{45}}+450, \newline C=450e^{\frac{1}{45}}.\newline \text{Therefore, } k(t)=450e^{\frac{1}{45}}e^{-\frac{t}{90}}+450.\newline \text{From here } k(2.1)=450e^{\frac{1}{45}}e^{-\frac{2.1}{90}}+450=450e^{-\frac{1}{900}}+450=899,5.\newline$

2)

$\frac{dy}{dx}=x^2y^2, y(-3)=1, y(-2.8)-?\newline \frac{dy}{dx}=_{|(-3,1)}=9\cdot 1=9.$

The equation of tanget line:

$y-y_1=k(x-x_1), \newline y-1=9(x+3),\newline y=9x+28.$

Therefore, $y(-2.8)=9\cdot(-2.8)+28=2.8.$

3)

$\frac{dy}{dx}=-\frac{x}{y^2}, y(-6)=-3.\newline y^2dy=-xdx,\newline \frac{y^3}{3}=-\frac{x^2}{2}+C,$ where C - is some constant.

If y(-6)=-3, then

$-27=-\frac{3}{2}\cdot 36+C,\newline C=27.\newline \text{So, } y=\sqrt[3]{27-\frac{3}{2}x^2.}$

4)

$\frac{dy}{dx}=3\cos{x}y, y(\pi/2)=-2.\newline \frac{dy}{y}=3\cos{x}dx,\newline \ln y=3\sin x+C, \text{where C is some constant.}\newline y=Ce^{3\sin x},\newline \text{If } y(\pi/2)=-2, \text{then}\newline -2=Ce^{3\sin{\frac{\pi}{2}}},\newline C=-\frac{2}{e^3}.\newline \text{So, } y=-\frac{2e^{3\sin x}}{e^3}.$