Question

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

1) given the differential equation

$\frac {dk}{dt}=\frac {-1}{90} (k-450); k(2)= 900 ...(i)$

is satisfied by thw function k(t) modeling the amount oof the substance; we can use the linear appriximation $k-k(a)= \frac {dk}{dt} (a)(t-a)...(ii)$

centered at [a,k(a)]

in this case, a=2 and k(a)=900.

$\therefore k-900 =\frac{dk}{dt} (2)[t-2]...(iii)$

from (i)$\frac {dk}{dt} (2)=\frac {-1}{90} [900-450]=-5 ...(iv)$

substituting (iv) in (iii)gives;

$k=900-5(t-2) or k=910-5t ...(v)$

as the linear approximation to k(t) satisfying (i)centered at (2, 900).

we can use (v) to approximate k(2.1)as;

$k(2.1) \approx 910-5(2.1)= 899.5$

2) for the differential equation $\frac {dy}{dx} =x^2y^2, y(-3)=1...(i)$ where y=f(x)

is the particular solution the tangent line equation at (-3,1) is the linear approximation to f(x) given by;

$y-y(-3)= \frac {dy}{dx} (-3,1)[x+3] or y-1 = \frac {dy}{dx} (-3,1)[x+3] ...(ii)$

from (i), $\frac {dy}{dx} (-3,1)= (-3)^2 (1) =9 ...(iii)$

substituting (iii) in (ii) gives the tangent equation to the curve [y=f(x) satisfying (i)] at (-3,1) to be

$y= 1+9(x+3) or y=9x+28 ...(iv)$

therefore, the approximate value of f(-2.8) is;

$y(-2.8)\approx f(-2.8) \approx 9(-2.8)+28 =2.8$

3) to solve

$\frac {dy}{dx} = \frac x{y^2}, y(-6)=-3 ...(i)$

we first separate the variables to get

$y^2 dy =xdx ...(ii)$

integrating (ii) we have;

$\frac {y^3}3= \frac {x^2}2+c ; y^3 = \frac32 x^2 +3c \implies y=\sqrt[3] {\frac 32 x^2 +k} for k=3c ...(iii)$

given that y(-6)= -3

we have

$-3= \sqrt[3] {\frac 32 (-6)^2 +k} ; -27=54+k \implies k=-81 ...(iv)$

substituting (iv) in (iii) gives the particular solution to (i) as;

$y= \sqrt [3] {\frac 32 x^2 -81}$

4) given; $\frac {dy}{dx} 3cos (x)y; y(\frac {\pi} 2)=-2 ...(i)$

then, separating the variables $\frac {dy}y =3cos (x)dx ...(ii)$

integrating (ii) we have $ln y= 3sinx +c; y= e^{3sinx+c}$

$y= Ae ^{3sinx} for A= e^c ...(iii)$

given $y(\frac {\pi} 2)=-2$ then,

$Ae^{3sin \frac {\pi}2}=-2 \implies A=-2e^{-3} ...(iv)$

substituting (iv) in (iii) gives the particular solution to (i) as

$y= -2e^{3sinx-3}$