Finding a Particular Solution Using Separation of Variables III Exercises 17-26, find the particular solution of the differential equation that satisfies the initial condition. Differential EquationInitial Condition y ( 1 + x 2 ) y ' − x ( 1 + y 2 ) = 0 y ( 0 ) = 3
Finding a Particular Solution Using Separation of Variables III Exercises 17-26, find the particular solution of the differential equation that satisfies the initial condition. Differential EquationInitial Condition y ( 1 + x 2 ) y ' − x ( 1 + y 2 ) = 0 y ( 0 ) = 3
Solution Summary: The author explains that the differential equation is in variable separable form, y(1+x2) and the integral formula.
Finding a Particular Solution Using Separation of Variables III Exercises 17-26, find the particular solution of the differential equation that satisfies the initial condition.
Differential EquationInitial Condition
y
(
1
+
x
2
)
y
'
−
x
(
1
+
y
2
)
=
0
y
(
0
)
=
3
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.
Tutorial Exercise
Consider the differential equation
x²y" - 4xy' + 6y=0; x², x³², (0, on).
Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.
Form the general solution.
Step 1
We are given the following homogenous differential equation and pair of solutions on the given interval.
x²y" - 4xy + 6y=0; x², x³, (0, 0)
We are asked to verify that the solutions are linearly independent. That is, there do not exist constants c, and c₂, not both zero, s
are different powers of x, we have a formal test to verify the linear independence.
Recall the definition of the Wronskian for the case of two functions f, and f, each of which have a first derivative.
f₂
W(f₁, f₂) =
Will HOUTBEave any pronTIES FOR LA
By Theorem 4.1.3, if W(f₁, f₂) 0 for every x in the interval of the solution, then solutions are linearly independent.
Let f(x) = x² and f(x) = x³. Complete the Wronskian for these functions.
x²
w(x², x³) =
2x
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