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logo1 Overview An Example Double Check Bernoulli Equations Bernd Schr ¨ oder Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Bernoulli Equations
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Overview An Example Double Check
Bernoulli Equations
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?
1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,where n 6= 0,1.
2. Recognizing Bernoulli equations requires some patternrecognition.
3. To solve a Bernoulli equation, we translate the equationinto a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting
linear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.
2. Recognizing Bernoulli equations requires some patternrecognition.
3. To solve a Bernoulli equation, we translate the equationinto a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting
linear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern
recognition.
3. To solve a Bernoulli equation, we translate the equationinto a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting
linear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern
recognition.3. To solve a Bernoulli equation, we translate the equation
into a linear equation.
3.1 The substitution y = v1
1−n turns the Bernoulli equationy′ +p(x)y = q(x)yn into a linear first order equation for v,
3.2 We can even write down the abstract form of the resultinglinear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern
recognition.3. To solve a Bernoulli equation, we translate the equation
into a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,
3.2 We can even write down the abstract form of the resultinglinear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern
recognition.3. To solve a Bernoulli equation, we translate the equation
into a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting
linear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern
recognition.3. To solve a Bernoulli equation, we translate the equation
into a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting
linear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,
where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern
recognition.3. To solve a Bernoulli equation, we translate the equation
into a linear equation.3.1 The substitution y = v
11−n turns the Bernoulli equation
y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting
linear first order equation, but it is simpler to remember thesubstitution y = v
11−n ,
3.3 After we solve the equation for v, we obtain y as theappropriate power of v.
That’s it.Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5,
y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5
= v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 ,
y′ =ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14
= −14
v−54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2y5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+ y = x2(
v−14
)5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
y′+(
v−14
)= x2
(v−
14
)5
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
−14
v−54 v′+
(v−
14
)= x2
(v−
14
)5
−14
v−54 v′+ v−
14 = x2v−
54
−14
v′+ v = x2
v′−4v = −4x2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
−14
v−54 v′+
(v−
14
)= x2
(v−
14
)5
−14
v−54 v′+ v−
14 = x2v−
54
−14
v′+ v = x2
v′−4v = −4x2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
−14
v−54 v′+
(v−
14
)= x2
(v−
14
)5
−14
v−54 v′+ v−
14 = x2v−
54
−14
v′+ v = x2
v′−4v = −4x2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Reduction to a linear equation.
n = 5, y = v1
1−5 = v−14 , y′ =
ddx
v−14 = −1
4v−
54 v′
−14
v−54 v′+
(v−
14
)= x2
(v−
14
)5
−14
v−54 v′+ v−
14 = x2v−
54
−14
v′+ v = x2
v′−4v = −4x2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x)
= e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx
= e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx
= e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x
(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation.
v′−4v = −4x2
µ(x) = e∫
p(x) dx = e∫−4 dx = e−4x
e−4x (v′−4v
)= −4x2e−4x
e−4xv′−4e−4xv = −4x2e−4x(e−4xv
)′= −4x2e−4x
e−4xv =∫−4x2e−4x dx
v = −4e4x∫
x2e−4x dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (continued).∫x2e−4x dx =
− 14
e−4xx2−∫ (
−14
)e−4x2x dx
= −14
e−4xx2 +12
∫e−4xx dx
= −14
e−4xx2 +12
[−1
4e−4xx−
∫−1
4e−4x dx
]= −1
4e−4xx2− 1
8e−4xx+
18
∫e−4x dx
= −14
e−4xx2− 18
e−4xx− 132
e−4x + c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (continued).∫x2e−4x dx = − 1
4e−4xx2−
∫ (−1
4
)e−4x2x dx
= −14
e−4xx2 +12
∫e−4xx dx
= −14
e−4xx2 +12
[−1
4e−4xx−
∫−1
4e−4x dx
]= −1
4e−4xx2− 1
8e−4xx+
18
∫e−4x dx
= −14
e−4xx2− 18
e−4xx− 132
e−4x + c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (continued).∫x2e−4x dx = − 1
4e−4xx2−
∫ (−1
4
)e−4x2x dx
= −14
e−4xx2 +12
∫e−4xx dx
= −14
e−4xx2 +12
[−1
4e−4xx−
∫−1
4e−4x dx
]= −1
4e−4xx2− 1
8e−4xx+
18
∫e−4x dx
= −14
e−4xx2− 18
e−4xx− 132
e−4x + c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (continued).∫x2e−4x dx = − 1
4e−4xx2−
∫ (−1
4
)e−4x2x dx
= −14
e−4xx2 +12
∫e−4xx dx
= −14
e−4xx2 +12
[−1
4e−4xx−
∫−1
4e−4x dx
]
= −14
e−4xx2− 18
e−4xx+18
∫e−4x dx
= −14
e−4xx2− 18
e−4xx− 132
e−4x + c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (continued).∫x2e−4x dx = − 1
4e−4xx2−
∫ (−1
4
)e−4x2x dx
= −14
e−4xx2 +12
∫e−4xx dx
= −14
e−4xx2 +12
[−1
4e−4xx−
∫−1
4e−4x dx
]= −1
4e−4xx2− 1
8e−4xx+
18
∫e−4x dx
= −14
e−4xx2− 18
e−4xx− 132
e−4x + c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (continued).∫x2e−4x dx = − 1
4e−4xx2−
∫ (−1
4
)e−4x2x dx
= −14
e−4xx2 +12
∫e−4xx dx
= −14
e−4xx2 +12
[−1
4e−4xx−
∫−1
4e−4x dx
]= −1
4e−4xx2− 1
8e−4xx+
18
∫e−4x dx
= −14
e−4xx2− 18
e−4xx− 132
e−4x + c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (concluded), general solution of theoriginal equation.
v = −4e4x[−1
4e−4xx2− 1
8e−4xx− 1
32e−4x + c
]= x2 +
12
x+18
+ ce4x
y = v−14 =
(x2 +
12
x+18
+ ce4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (concluded), general solution of theoriginal equation.
v = −4e4x[−1
4e−4xx2− 1
8e−4xx− 1
32e−4x + c
]
= x2 +12
x+18
+ ce4x
y = v−14 =
(x2 +
12
x+18
+ ce4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (concluded), general solution of theoriginal equation.
v = −4e4x[−1
4e−4xx2− 1
8e−4xx− 1
32e−4x + c
]= x2 +
12
x+18
+ ce4x
y = v−14 =
(x2 +
12
x+18
+ ce4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (concluded), general solution of theoriginal equation.
v = −4e4x[−1
4e−4xx2− 1
8e−4xx− 1
32e−4x + c
]= x2 +
12
x+18
+ ce4x
y = v−14
=(
x2 +12
x+18
+ ce4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Solving the linear equation (concluded), general solution of theoriginal equation.
v = −4e4x[−1
4e−4xx2− 1
8e−4xx− 1
32e−4x + c
]= x2 +
12
x+18
+ ce4x
y = v−14 =
(x2 +
12
x+18
+ ce4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.
y =(
x2 +12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0)
=(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78,
y =(
x2 +12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
Finding c.y =
(x2 +
12
x+18
+ ce4x)− 1
4
1 = y(0) =(
02 +12·0+
18
+ ce4·0)− 1
4
1 =(
18
+ c)− 1
4
1 =18
+ c
c =78, y =
(x2 +
12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.
y =(
x2 +12
x+18
+78
e4x)− 1
4
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
ddx
[(x2 +
12
x+18
+78
e4x)− 1
4]
+(
x2 +12
x+18
+78
e4x)− 1
4
= −14
(x2 +
12
x+18
+78
e4x)− 5
4(
2x+12
+72
e4x)
+(
x2 +12
x+18
+78
e4x)− 1
4
=(
x2 +12
x+18
+78
e4x)− 5
4(− x
2− 1
8− 7
8e4x + x2 +
12
x+18
+78
e4x)
=(
x2 +12
x+18
+78
e4x)− 5
4x2 = y5x2 √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
ddx
[(x2 +
12
x+18
+78
e4x)− 1
4]
+(
x2 +12
x+18
+78
e4x)− 1
4
= −14
(x2 +
12
x+18
+78
e4x)− 5
4(
2x+12
+72
e4x)
+(
x2 +12
x+18
+78
e4x)− 1
4
=(
x2 +12
x+18
+78
e4x)− 5
4(− x
2− 1
8− 7
8e4x + x2 +
12
x+18
+78
e4x)
=(
x2 +12
x+18
+78
e4x)− 5
4x2 = y5x2 √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
ddx
[(x2 +
12
x+18
+78
e4x)− 1
4]
+(
x2 +12
x+18
+78
e4x)− 1
4
= −14
(x2 +
12
x+18
+78
e4x)− 5
4(
2x+12
+72
e4x)
+(
x2 +12
x+18
+78
e4x)− 1
4
=(
x2 +12
x+18
+78
e4x)− 5
4(− x
2− 1
8− 7
8e4x + x2 +
12
x+18
+78
e4x)
=(
x2 +12
x+18
+78
e4x)− 5
4x2 = y5x2 √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
ddx
[(x2 +
12
x+18
+78
e4x)− 1
4]
+(
x2 +12
x+18
+78
e4x)− 1
4
= −14
(x2 +
12
x+18
+78
e4x)− 5
4(
2x+12
+72
e4x)
+(
x2 +12
x+18
+78
e4x)− 1
4
=(
x2 +12
x+18
+78
e4x)− 5
4(− x
2− 1
8− 7
8e4x + x2 +
12
x+18
+78
e4x)
=(
x2 +12
x+18
+78
e4x)− 5
4x2
= y5x2 √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
ddx
[(x2 +
12
x+18
+78
e4x)− 1
4]
+(
x2 +12
x+18
+78
e4x)− 1
4
= −14
(x2 +
12
x+18
+78
e4x)− 5
4(
2x+12
+72
e4x)
+(
x2 +12
x+18
+78
e4x)− 1
4
=(
x2 +12
x+18
+78
e4x)− 5
4(− x
2− 1
8− 7
8e4x + x2 +
12
x+18
+78
e4x)
=(
x2 +12
x+18
+78
e4x)− 5
4x2 = y5x2
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
ddx
[(x2 +
12
x+18
+78
e4x)− 1
4]
+(
x2 +12
x+18
+78
e4x)− 1
4
= −14
(x2 +
12
x+18
+78
e4x)− 5
4(
2x+12
+72
e4x)
+(
x2 +12
x+18
+78
e4x)− 1
4
=(
x2 +12
x+18
+78
e4x)− 5
4(− x
2− 1
8− 7
8e4x + x2 +
12
x+18
+78
e4x)
=(
x2 +12
x+18
+78
e4x)− 5
4x2 = y5x2 √
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
y(0) =(
02 +12·0+
18
+78
e4·0)− 1
4
=(
18
+78
)− 14
= 1√
Yes, it does.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
y(0) =(
02 +12·0+
18
+78
e4·0)− 1
4
=(
18
+78
)− 14
= 1√
Yes, it does.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
y(0) =(
02 +12·0+
18
+78
e4·0)− 1
4
=(
18
+78
)− 14
= 1
√
Yes, it does.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
y(0) =(
02 +12·0+
18
+78
e4·0)− 1
4
=(
18
+78
)− 14
= 1√
Yes, it does.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
logo1
Overview An Example Double Check
Does y =(
x2 +12
x+18
+78
e4x)−1
4
Really Solve
the Initial Value Problem y′+ y = x2y5, y(0) = 1?
y(0) =(
02 +12·0+
18
+78
e4·0)− 1
4
=(
18
+78
)− 14
= 1√
Yes, it does.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Bernoulli Equations
