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9
2010 Calculus #9: Trigonometric Differential Equation
2010 Calculus #9: Trigonometric Differential Equation
Source:
July 15, 2012
calculus
trigonometry
Problem Statement
Let
x
(
t
)
x(t)
x
(
t
)
be a solution to the differential equation
(
x
+
x
′
)
2
+
x
⋅
x
′
′
=
cos
t
\left(x+x^\prime\right)^2+x\cdot x^{\prime\prime}=\cos t
(
x
+
x
′
)
2
+
x
⋅
x
′′
=
cos
t
with
x
(
0
)
=
x
′
(
0
)
=
2
5
x(0)=x^\prime(0)=\sqrt{\frac{2}{5}}
x
(
0
)
=
x
′
(
0
)
=
5
2
. Compute
x
(
π
4
)
x\left(\dfrac{\pi}{4}\right)
x
(
4
π
)
.
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