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.
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