Explanation Given: The differential equation is x 2 y ″ + x y ′ + y = sec ( ln x ) . Definition: “A solution in which the dependent variable is expressed solely in terms of the independent variable and constant is said to be an explicit solution.” Calculation: Consider the function y = cos ( ln x ) ln ( cos ( ln x ) ) + ( ln x ) sin ( ln x ) . Differentiate the given function with respect to x . d d x y = d d x [ cos ( ln x ) ln ( cos ( ln x ) ) + ( ln x ) sin ( ln x ) ] y ′ = cos x ( ln x ) − 1 x ( sin ln x ) cos x ( ln x ) − sin ( ln x ) x ln ( cos ( ln x ) ) + sin ( ln x ) x + ( ln x ) cos ( ln x ) x = ( ln x ) cos ( ln x ) − sin ( ln x ) ln cos ( ln x ) x ( 1 ) Differentiate again with respect to x . y ″ = 1 x 2 { x [ cos ( ln x ) x − ( ln x ) sin ( ln x ) x − cos ( ln x ) x ln cos ( ln x ) + sin ( ln x ) sin ( ln x ) x cos ( ln x ) ] − [ ( ln x ) cos ( ln x ) − sin ( ln x ) ln cos ( ln x ) ] } = 1 x 2 { cos<

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ISBN: 9781337292405

Author: ZILL

Publisher: CENGAGE L

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 1, Problem 26RE

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To verify: The function is an explicit solution of the given differential equation and to find an interval of definition.

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Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist. 8+x f(x) = x(x-1) (Use a comma to separate answers as needed.) OA. The function f is discontinuous at the single value x = OB. The function f is discontinuous at the single value x = OC. The function f is discontinuous at the two values x = OD. The function f is discontinuous at the two values x = not oo or -0. OE. The function f is discontinuous at the two values x = The limit is The limit does not exist and is not oo or - co. The limits for both values do not exist and are not co or - co. The limit for the smaller value is The limit for the larger value does not exist and is The limit for the smaller value does not exist and is not co or - co. The limit for the larger

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Q1 4 Points In each part, determine if the given set H is a subspace of the given vector space V. Q1.1 1 Point V = R and H is the set of all vectors in R4 which have the form a b x= 1-2a for some scalars a, b. 1+3b 2 (e.g., the vector x = is an example of one element of H.) OH is a subspace of V. OH is not a subspace of V. Save Answer Q1.2 1 Point V = P3, the vector space of all polynomials whose degree is at most 3, and H = +³, 3t2}. OH is a subspace of V. OH is not a subspace of V. Save Answer Span{2+ Q1.3 1 Point V = M2x2, the vector space of all 2 x 2 matrices, and H is the subset of M2x2 consisting of all invertible 2 × 2 matrices. OH is a subspace of V. OH is not a subspace of V. Save Answer

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To verify: The function is an explicit solution of the given differential equation and to find an interval of definition.

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To verify: The function is an explicit solution of the given differential equation and to find an interval of definition.

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Chapter 1 Solutions