The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0

3 years ago

How do you right 0.0001 in expanded form

kirza4 [7]

The expanded form of the number 0.0001 is $\bold{\frac{1}{10000}}$ .

<u>SOLUTION: </u>

Given that, we have to write 0.0001 in expanded form.

Expanded form or expanded notation is a way of writing numbers to see the math value of individual digits. When numbers are separated into individual place values and decimal places they can also form a mathematical expression.

Now, take the given number 0.0001

As there are no other digits except 0 in front of 1 our work is simplified.

Expanded form will be $\frac{1}{10000}$

Hence, the expanded form of the number 0.0001 is $\frac{1}{10000}$ .

4 0

3 years ago