Answer:
So, solution of the differential equation is
Step-by-step explanation:
We have the given differential equation: y′′+4y=5xcos(2x)
We use the Method of Undetermined Coefficients.
We first solve the homogeneous differential equation y′′+4y=0.
It is a homogeneous solution:
Now, we finding a particular solution.
we get
So, solution of the differential equation is
Answer:
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4
Step-by-step explanation:
The given function is presented as follows;
From the given function, we have;
When x = 1, the denominator of the fraction, , which is (x - 1) = 0, and the function becomes,
therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote
Also we have that in the given function, as <em>x</em> increases, the fraction tends to 0, therefore as x increases, we have;
Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1
When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, . When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left
Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.