Calculus Question (x2) Involving Differential Equations?

dx/dt = 3/x

separate the variables

3 x dx = 3 dt

Integrate both sides

3 ∫ x dx = 3 ∫ dt

(3) (1/2)x^2 = 3t + C

(3/2) x^2 = 3t + C

x(0) = 5

(3/2)(5)^2 = 3(0) + C

C = 75/2

(3/2) x^2 = 3t + 75/2

multiply both sides by 2/3

x^2 = 6t + 25

x= sqrt(6t+25)

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dy/dx + y/4 = 0

dy/dx = -y/4

separate the variables

dy/y = -4 dx

Integrate both sides

∫ dy/y = -(1/4) ∫ dx

ln(y) = (-1/4)x + C1

y = e^(-x/4) e^C1

y = C e^(-x/4)

y(0) = 14

y=14; x=0

14 = C e^0

C = 14

y = 14 e^(-x/4)

There's a neat method that nobody who's answered so far seems to use that "bakes in" the initial value, without needing a separate "solve for C" step.

In the first problem, dx/dt = 3/x can be written in separated form:

x dx = 3 dt

Think of the solution as a curve on an x-vs-t graph, Lots of curves work, but you're trying to find the one that passes through the point (t_0, x_0) = (0, 5). Take *definite* integrals:

∫ x dx = ∫ 3 dt .... with limits x=x_0 to x on the left and t=t_0 to t on the right.

x²/2 - 5²/2 = 3t - 0

x² - 25 = 6t

x = √(6t + 25) .... choosing the + root to make (0, 5) work at t=0.

The same approach works on the second problem, but this time I'll wait to plug in those initial values:

dy/dx = -y/4

(1/y) dy = (1/4) dx

Integrate that from y=y_0 to y on the left, x=x_0 to x on the right:

log |y| - log |y_0| = (x - x_0)/4

log |y / y_0| = (x - x_0)/4

Note that this solution can't have y=0 at any point, so the curve of a solution can't cross the x-axis. That means y and y_0 have the same sign and the absolute value bars can be dropped. Then take exponentials and:

y = y_0 * e^[(x - x_0) / 4]

That's the general initial value solution. Get your answer by substituting (x_0, y_0) = (0, 14):

y = 14 e^(x/4)

1) Cauchy problem

dx/dt=3/x ; with initial condition x(0)=5

1.1 Separate the variables

xdx=3dt

1.2 Integrate both sides

∫x dx=3∫dt

x²/2=3t+C ; C is a generic real constant

x²=6t+k ; k is a generic real constant (we can call it also C)

x=±√(6t+C)

1.3 Initial condition

x(0)=±√(0+C)=5

C=+5²=25

hence the solution of Cauchy problem is x(t)=±√(6t+25)

2. Cauchy problem

dy/dx+y/4=0 ; with initial condition y(0)=14

2.1 Separate the variables

dy/dx=-y/4

dy/y=-4dx

2.2 Integrate both sides

∫1/y dy=-4∫dx

ln|y|=-4x+C

e^ln|y|=e^(-4x+C)=e^C*e⁻⁴ˣ

where C is a generic real constant.

|y|=e^C/e⁴ˣ

hence, two solutions

y=e^C/e⁴ˣ

y=-e^C/e⁴ˣ

2.3 Cauchy problem y(0)=14 is refers to the positive solution,

y=e^C/e⁴ˣ

y(0)=e^C/1=14

The unique solution of our problem is

y(x)=14/e⁴ˣ

1) dx/dt = 3/x

xdx = 3dt ...........separating variables

∫ xdx = 3∫dt

x^2/2 = 3t + C

For t = 0. x = 5

25/2 = C

x^2/2 = 3t + 25/2

x^2 = 6t + 25

x(t) = ±√(6t + 25)

2) (dy/dx) + (y/4) = 0

dy/dx = - y/4

dy/y = -1/4dx ...........separating variables

∫(1/y) dy = (-1/4) ∫dx

ln(y) = -x/4 + C

For x = 0, y = 14

ln(14) = C

ln(y) = -x/4 + ln(14)

ln(y) - ln(14) = -x/4

ln(y/14) = - x/4

y/14 = e^(-x/4)

y = 14 e^(-x/4)

Where is the original question.

(dx/dt) = (3/x) , x(0) = 5

x dx = 3 dt --> x²/2 = 3t + C. The condition says 5²/2 = 3(0) + C thus C = ...

Next one, (dy/dx) + (y/4) = 0 , y(0) = 14

dy/dx = -y/4 --> dy/y = (-1/4) dx --> ln(y) = -x/4 + C ... You can now take it from there.

Done!

Lol cool bro