Given that 1 + t , 1 + 2 t , and 1 + 3 t 2 are solutions to the differential equation y ′′ + p ( t ) y ′ + q ( t ) y = g ( t ) , find the solution to this equation that satisfies y ( 1 ) = 2 , y ′ ( 1 ) = 0 .
Given that 1 + t , 1 + 2 t , and 1 + 3 t 2 are solutions to the differential equation y ′′ + p ( t ) y ′ + q ( t ) y = g ( t ) , find the solution to this equation that satisfies y ( 1 ) = 2 , y ′ ( 1 ) = 0 .
Solution Summary: The author explains that the solution to the given differential equation that satisfies the initial conditions is 2t-t2+1.
Given that
1
+
t
,
1
+
2
t
, and
1
+
3
t
2
are solutions to the differential equation
y
′′
+
p
(
t
)
y
′
+
q
(
t
)
y
=
g
(
t
)
, find the solution to this equation that satisfies
y
(
1
)
=
2
,
y
′
(
1
)
=
0
.
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01 - What Is A Differential Equation in Calculus? Learn to Solve Ordinary Differential Equations.; Author: Math and Science;https://www.youtube.com/watch?v=K80YEHQpx9g;License: Standard YouTube License, CC-BY
Higher Order Differential Equation with constant coefficient (GATE) (Part 1) l GATE 2018; Author: GATE Lectures by Dishank;https://www.youtube.com/watch?v=ODxP7BbqAjA;License: Standard YouTube License, CC-BY