For Problems 28–32, show that the given relation defines an implicit solution to the given differential equation, where c is an arbitrary constant. x y 2 + 2 y − x = c , y ′ = 1 − y 2 2 ( 1 + x y ) .
For Problems 28–32, show that the given relation defines an implicit solution to the given differential equation, where c is an arbitrary constant. x y 2 + 2 y − x = c , y ′ = 1 − y 2 2 ( 1 + x y ) .
Solution Summary: The author explains that the relation y2+2y-x=c defines an implicit solution to the differential equation.
For Problems 28–32, show that the given relation defines an implicit solution to the given differential equation, where c is an arbitrary constant.
x
y
2
+
2
y
−
x
=
c
,
y
′
=
1
−
y
2
2
(
1
+
x
y
)
.
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
For each dif erential equation in Problems 1–21, find the general solutionby finding the homogeneous solution and a particular solution.
Please DO NOT YOU THE PI method where 1/f(r) * x. Dont do that.
Instead do this, assume for yp = to something, do the 1 and 2 derivative of it and then plug it in the equation to find the answer.
The instructions say:
For each differential equation in Problems 1–21, find the general solution by finding the homogeneous solution and a particular solution.
The first image below is the problem, the second is the answer. I'm able to get the homogeneous solution but am struggling with getting the particular solution to get to the answer the textbook provides.
C. Eliminate the arbitrary constants in each equation and express the final answers in the
following form: a,n(x)y(n) + an-1(x)yn-1) + . .. + a1(x)y' + ao(x)y – g(x) = 0 .
1. r* – y² = cy
2. y = cje" + cze²" + c3e*r
Chapter 1 Solutions
Differential Equations and Linear Algebra (4th Edition)
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