What is the solution for the differential equation: 2y' = y?

Answer 1

Rewrite the equation so it looks like

2y'-y=0.

This is a first order linear homogeneous equation with constant coefficients so its solution is determined by its characteristic equation:

2k-1=0 -> k=1/2

so the general solution is y=C e^(t/2).

Of course, you could also solve it by multiplying through by an integrating factor:

e^(-t/2) y'-1/2 e^(-t/2)y=0

d/dt( e^(-t/2) y) =0

e^(-t/2) y=C

y=Ce^(t/2)

Or, you could solve it by noting that it is separable, too:

1/y dy= 1/2 dt

ln |y|=1/2 t+C

y=Ce^(t/2)

Answer 2

1=2

Answer 3

Y=0

Answer 4

y=e^(x/2) => y'=1/2 e^(x/2) = y/2

Answer 5

let P S y_p = ax^2 + bx + c + dcos(x) + esin(x) Find dy_p/dx and d^2y_p/dx2 and pluf them into the ODE and then compare coefficients with 1+(x^2)+2cos(x) to get y_p = x^2 + 4x + 7 -sin(x) therefore y = (Ax + B)e^x + x^2 + 4x + 7 -sin(x) Substitute the initial conditions to get A = 1 and B = -1 which gives the general solution y = (x -1)e^x + x^2 + 4x + 7 -sin(x)

Answer 6

y=e^(x/2)

Answer 7

Rewrite y' as dy/dx
2(dy/dx) = y
Multiply both sides by dx
2(dy) = y(dx)
Divide both sides by y
(2/y)dy = dx
Now you can see that this is a separable equation.
Integrate both sides.
2ln(y) = x + k
Divide both sides by 2.
ln(y) = x/2 + k/2
Rename the constant k/2 as c.
ln(y) = x/2 + c
Apply the exponential, e, to both sides.
y = e^(x/2 +c)
y = e^(x/2) e^c
Since e^c is just a constant you can relabel it as C.
Answer: y = Ce^(x/2)

Answer 8

I think y=0

Answer 9

the derivative of e^f(x) = f'(x)e^f(x)
therefore by making f(x) = 0.5x the derivative of the function will be half the value of the original function:
y = e^(0.5x)
y' = 0.5*e^(0.5x)
2y' = 2*0.5*e^(0.5x)
2y' = e^0.5x

:-)

Answer 10

2y' = y

2*(dy/dx) = y

dx / 2 = dy / y

x / 2 + C = ln(y)
e^(x/2 + C) = y

C*e^(x/2) = y