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Let's here start thinking regarding that how to solve nonhomogeneous differential equations. A second order, linear non-homogeneous differential equation is as
y′′ + p (t) y′ +q (t) y = g (t ) .....................(1)
Here g(t) is a non-zero function. Note that we didn't go along with constant coefficients here since everything that we're going to do under this section doesn't need it. Also, we're using a coefficient of 1 on the second derivative just to create some of the work a little simple to write down. This is not needed to be a 1.
Before talking about how to resolve one of these we require to get some fundamentals out of the way that are the point of this section.
First, we will call
y′′ + p (t ) y′ + q (t ) y = 0 (2)
It is the associated homogeneous differential equation to (1). Here, let's take a look at the subsequent theorem.
Find out the volume of the solid obtained by rotating the region bounded by y = x 2 - 2x and y = x about the line y = 4 . Solution: Firstly let's get the bounding region & t
56+3
log4 (2x+4)-3=log4 3
how do you turn 91 divided by730 into a compatible number
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We are until now going to suppose that there will be no external forces acting on the system, along with the exception of damping obviously. Under this case the differential equati
A rectangular container is 15 cm wide and 5 cm long, and contains water to a depth of 8 cm. An object is placed in the water and the water rises 2.3 cm. Determine the volume of the
1. Which of the following is greater than 4.3 x 10^9 a. 2.1 x 10^9 b. 3.2 x 10^9 c. 5.3 x 10^9 d. 7.4 x 10^8 2. Which of the following is less than 6.5 x 10^-5 a. 1.4 x 10
calculates the value of the following limit. Solution Now, notice that if we plug in θ =0 which we will get division by zero & so the function doesn't present at this
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