Solving Initial Value Problems
An Example
Double Check
An Initial Value Problem for a Separable
Differential Equation
Bernd Schröder
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
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Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Approach
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Approach
This approach will work as long as the general solution of the
differential equation can be computed.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Approach
This approach will work as long as the general solution of the
differential equation can be computed.
1. Find the general solution of the differential equation.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Approach
This approach will work as long as the general solution of the
differential equation can be computed.
1. Find the general solution of the differential equation.
2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Approach
This approach will work as long as the general solution of the
differential equation can be computed.
1. Find the general solution of the differential equation.
2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.
3. Double check if the solution works.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Approach
This approach will work as long as the general solution of the
differential equation can be computed.
1. Find the general solution of the differential equation.
2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.
3. Double check if the solution works.
That’s it.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solve the Initial Value Problem
x sin(x)
y0 = x sin(x)y +
, y(0) = 1.
y
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solve the Initial Value Problem
x sin(x)
y0 = x sin(x)y +
, y(0) = 1.
y
y0 = x sin(x)y +
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
x sin(x)
y
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solve the Initial Value Problem
x sin(x)
y0 = x sin(x)y +
, y(0) = 1.
y
x sin(x)
y0 = x sin(x)y +
y
1
0
y = x sin(x) y +
y
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solve the Initial Value Problem
x sin(x)
y0 = x sin(x)y +
, y(0) = 1.
y
x sin(x)
y0 = x sin(x)y +
y
1
0
y = x sin(x) y +
y
2
y +1
dy
= x sin(x)
dx
y
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solve the Initial Value Problem
x sin(x)
y0 = x sin(x)y +
, y(0) = 1.
y
x sin(x)
y0 = x sin(x)y +
y
1
0
y = x sin(x) y +
y
2
y +1
dy
= x sin(x)
dx
y
y
dy = x sin(x) dx
y2 + 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solve the Initial Value Problem
x sin(x)
y0 = x sin(x)y +
, y(0) = 1.
y
x sin(x)
y0 = x sin(x)y +
y
1
0
y = x sin(x) y +
y
2
y +1
dy
= x sin(x)
dx
y
Z
y
dy =
y2 + 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
Z
x sin(x) dx
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Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
du
du
= 2y, so dy = .
dy
2y
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
du
du
= 2y, so dy = .
dy
2y
Z
y
dy =
2
y +1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
du
du
= 2y, so dy = .
dy
2y
Z
Z
y
y du
dy =
2
u 2y
y +1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
du
du
= 2y, so dy = .
dy
2y
Z
Z
y
y du
dy =
2
u 2y
y +1
Z
1 1
=
du
2 u
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
du
du
= 2y, so dy = .
dy
2y
Z
Z
y
y du
dy =
2
u 2y
y +1
Z
1 1
=
du
2 u
1
=
ln |u| + c
2
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Z
The Integral
Double Check
y
dy.
y2 + 1
Idea: Substitution u := y2 + 1.
du
du
= 2y, so dy = .
dy
2y
Z
Z
y
y du
dy =
2
u 2y
y +1
Z
1 1
=
du
2 u
1
=
ln |u| + c
2
1 2
=
ln y + 1 + c
2
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Z
x sin(x) dx =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Z
x sin(x) dx = x − cos(x) −
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Z
Z
x sin(x) dx = x − cos(x) − 1 · − cos(x) dx
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Z
Z
x sin(x) dx = x − cos(x) − 1 · − cos(x) dx
= −x cos(x) +
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
Z
cos(x) dx
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Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Z
Z
x sin(x) dx = x − cos(x) − 1 · − cos(x) dx
= −x cos(x) +
Z
cos(x) dx
= −x cos(x) + sin(x) + c
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Z
The Integral
x sin(x) dx.
Idea: Integration by parts.
Integrate sin(x), differentiate x.
Z
Z
x sin(x) dx = x − cos(x) − 1 · − cos(x) dx
= −x cos(x) +
Z
cos(x) dx
= −x cos(x) + sin(x) + c
= sin(x) − x cos(x) + c
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
Z
y
dy =
2
y +1
Z
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
x sin(x) dx
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
2
y +1
1 2
ln y + 1 =
2
Z
Z
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
x sin(x) dx
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
x sin(x) dx
2
y +1
1 2
ln y + 1 = sin(x) − x cos(x) + c
2
Z
Z
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
x sin(x) dx
2
y +1
1 2
ln y + 1 = sin(x) − x cos(x) + c
2 ln y2 + 1 = 2 sin(x) − 2x cos(x) + c
Z
Z
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
x sin(x) dx
2
y +1
1 2
ln y + 1 = sin(x) − x cos(x) + c
2 ln y2 + 1 = 2 sin(x) − 2x cos(x) + c
Z
Z
y2 + 1 = e2 sin(x)−2x cos(x)+c
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
x sin(x) dx
2
y +1
1 2
ln y + 1 = sin(x) − x cos(x) + c
2 ln y2 + 1 = 2 sin(x) − 2x cos(x) + c
Z
Z
y2 + 1 = ke2 sin(x)−2x cos(x)
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
x sin(x) dx
2
y +1
1 2
ln y + 1 = sin(x) − x cos(x) + c
2 ln y2 + 1 = 2 sin(x) − 2x cos(x) + c
Z
Z
y2 + 1 = ke2 sin(x)−2x cos(x)
y2 = ke2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
... Continuing Where We Left Off.
y
dy =
x sin(x) dx
2
y +1
1 2
ln y + 1 = sin(x) − x cos(x) + c
2 ln y2 + 1 = 2 sin(x) − 2x cos(x) + c
Z
Z
y2 + 1 = ke2 sin(x)−2x cos(x)
y2 = ke2 sin(x)−2x cos(x) − 1
p
y = ± ke2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
1 =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
p
1 = ± ke0 − 1 =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
p
√
1 = ± ke0 − 1 = ± k − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
p
√
1 = ± ke0 − 1 = ± k − 1
√
1 =
k−1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
p
√
1 = ± ke0 − 1 = ± k − 1
√
1 =
k−1
1 = k−1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
p
√
1 = ± ke0 − 1 = ± k − 1
√
1 =
k−1
1 = k−1
k = 2
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
1 = y(0)
p
y(x) = ± ke2 sin(x)−2x cos(x) − 1
p
1 = ± ke2 sin(0)−2·0·cos(0) − 1
p
√
1 = ± ke0 − 1 = ± k − 1
√
1 =
k−1
1 = k−1
k = 2
y(x) =
p
2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
y(x) =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
p
2e2 sin(x)−2x cos(x) − 1
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
Solving the Initial Value Problem
y(x) =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
p
2e2 sin(x)−2x cos(x) − 1
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
y(x) =
p
2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
p
2e2 sin(x)−2x cos(x) − 1
p
y(0) =
2e2 sin(0)−2·0·cos(0) − 1
y(x) =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
p
2e2 sin(x)−2x cos(x) − 1
p
p
y(0) =
2e2 sin(0)−2·0·cos(0) − 1 = 2e0 − 1
y(x) =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
p
2e2 sin(x)−2x cos(x) − 1
p
p
y(0) =
2e2 sin(0)−2·0·cos(0) − 1 = 2e0 − 1 = 1
y(x) =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
p
2e2 sin(x)−2x cos(x) − 1
p
p
y(0) =
2e2 sin(0)−2·0·cos(0) − 1 = 2e0 − 1 = 1
y(x) =
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
√
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
1
= p
2 2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
1
= p
2e2 sin(x)−2x cos(x)
2
sin(x)−2x
cos(x)
2 2e
−1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
2e2 sin(x)−2x cos(x)
= p
2 2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
2e2 sin(x)−2x cos(x)
2 cos(x) − 2 cos(x) + 2x sin(x)
= p
2 2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
2e2 sin(x)−2x cos(x)
2 cos(x) − 2 cos(x) + 2x sin(x)
= p
2 2e2 sin(x)−2x cos(x) − 1
2e2 sin(x)−2x cos(x)
= p
2x sin(x)
2 2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
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Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
p
Does y(x) = 2e2 sin(x)−2x cos(x) − 1 Solve the IVP
x sin(x)
y0 = x sin(x)y +
, y(0) = 1?
y
d p 2 sin(x)−2x cos(x)
2e
−1
dx
2e2 sin(x)−2x cos(x)
2 cos(x) − 2 cos(x) + 2x sin(x)
= p
2 2e2 sin(x)−2x cos(x) − 1
2e2 sin(x)−2x cos(x)
= p
2x sin(x)
2 2e2 sin(x)−2x cos(x) − 1
2e2 sin(x)−2x cos(x)
= p
x sin(x)
2e2 sin(x)−2x cos(x) − 1
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
x sin(x)y +
An Example
Double Check
x sin(x)
y
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
x sin(x)
y
p
x sin(x)
= x sin(x) 2e2 sin(x)−2x cos(x) − 1 + p
2e2 sin(x)−2x cos(x) − 1
x sin(x)y +
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
x sin(x)
y
p
x sin(x)
= x sin(x) 2e2 sin(x)−2x cos(x) − 1 + p
2e2 sin(x)−2x cos(x) − 1
p
2
x sin(x)
2e2 sin(x)−2x cos(x) − 1 + x sin(x)
p
=
2e2 sin(x)−2x cos(x) − 1
x sin(x)y +
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
x sin(x)
y
p
x sin(x)
= x sin(x) 2e2 sin(x)−2x cos(x) − 1 + p
2e2 sin(x)−2x cos(x) − 1
p
2
x sin(x)
2e2 sin(x)−2x cos(x) − 1 + x sin(x)
p
=
2e2 sin(x)−2x cos(x) − 1
x sin(x) 2e2 sin(x)−2x cos(x) − 1 + 1
p
=
2e2 sin(x)−2x cos(x) − 1
x sin(x)y +
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
x sin(x)
y
p
x sin(x)
x sin(x) 2e2 sin(x)−2x cos(x) − 1 + p
2e2 sin(x)−2x cos(x) − 1
p
2
x sin(x)
2e2 sin(x)−2x cos(x) − 1 + x sin(x)
p
2e2 sin(x)−2x cos(x) − 1
x sin(x) 2e2 sin(x)−2x cos(x) − 1 + 1
p
2e2 sin(x)−2x cos(x) − 1
2e2 sin(x)−2x cos(x)
x sin(x) p
2e2 sin(x)−2x cos(x) − 1
x sin(x)y +
=
=
=
=
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science
Solving Initial Value Problems
An Example
Double Check
x sin(x)
y
p
x sin(x)
x sin(x) 2e2 sin(x)−2x cos(x) − 1 + p
2e2 sin(x)−2x cos(x) − 1
p
2
x sin(x)
2e2 sin(x)−2x cos(x) − 1 + x sin(x)
p
2e2 sin(x)−2x cos(x) − 1
x sin(x) 2e2 sin(x)−2x cos(x) − 1 + 1
p
2e2 sin(x)−2x cos(x) − 1
√
2e2 sin(x)−2x cos(x)
x sin(x) p
2e2 sin(x)−2x cos(x) − 1
x sin(x)y +
=
=
=
=
Bernd Schröder
An Initial Value Problem for a Separable Differential Equation
logo1
Louisiana Tech University, College of Engineering and Science