A Second Order Ordinary Differential equation with one known solution
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This problem is from the book "Introduction to Ordinary Differential Equations" by Shepley L. Ross. I am thinking I did the problem correctly because my answer matches the answer in the back of the book. What bothers me is the fact the second solution includes the first solution.
Problem:
Given that $y = x$ is a solution of
$$ x^2y'' - 4xy' + 4y = 0$$
find a linearly independent solution by reducing the order. Write the general solution.
Answer:
Let $f(x)$ represent the solution we have.
begin{eqnarray*}
f(x) &=& x \
y &=& f(x) v = xv \
y' &=& x v' + v \
y'' &=& x v'' + v' + v' = xv'' + 2v' \
end{eqnarray*}
begin{eqnarray*}
x^2(xv'' + 2v') - 4x(x v' + v) + 4xv &=& 0 \
x^3v'' + 2x^2v' - 4x^2v' &=& 0 \
x^3v'' - 2x^2v' &=& 0 \
text{Let }w &=& frac{dv}{dx} \
x^3 w' - 2x^2 w &=& 0 \
x w' - 2 w &=& 0 \
x w' &=& 2w \
x frac{dw}{dx} &=& 2w \
frac{x}{dx} &=& frac{2w}{dw} \
frac{x}{2dx} &=& frac{w}{dw} \
frac{ 2dx}{x} &=& frac{dw}{w} \
2ln{x} &=& ln{w} + c_0 \
x^2 &=& c_1 w \
x^2 &=& c_1 frac{dv}{dx} \
frac{dv}{dx} &=& c_2 x^2 \
v &=& c_3 x^3 + c_4 \
frac{y}{x} &=& c_3 x^3 + c_4 \
y &=& c_3 x^4 + c_4 x \
end{eqnarray*}
Hence the general solution is:
$$ y = C_0x^4 + C_1x $$
integration ordinary-differential-equations
asked Dec 2 '18 at 2:50
BobBob
918515
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add a comment |
$begingroup$
This problem is from the book "Introduction to Ordinary Differential Equations" by Shepley L. Ross. I am thinking I did the problem correctly because my answer matches the answer in the back of the book. What bothers me is the fact the second solution includes the first solution.
Problem:
Given that $y = x$ is a solution of
$$ x^2y'' - 4xy' + 4y = 0$$
find a linearly independent solution by reducing the order. Write the general solution.
Answer:
Let $f(x)$ represent the solution we have.
begin{eqnarray*}
f(x) &=& x \
y &=& f(x) v = xv \
y' &=& x v' + v \
y'' &=& x v'' + v' + v' = xv'' + 2v' \
end{eqnarray*}
begin{eqnarray*}
x^2(xv'' + 2v') - 4x(x v' + v) + 4xv &=& 0 \
x^3v'' + 2x^2v' - 4x^2v' &=& 0 \
x^3v'' - 2x^2v' &=& 0 \
text{Let }w &=& frac{dv}{dx} \
x^3 w' - 2x^2 w &=& 0 \
x w' - 2 w &=& 0 \
x w' &=& 2w \
x frac{dw}{dx} &=& 2w \
frac{x}{dx} &=& frac{2w}{dw} \
frac{x}{2dx} &=& frac{w}{dw} \
frac{ 2dx}{x} &=& frac{dw}{w} \
2ln{x} &=& ln{w} + c_0 \
x^2 &=& c_1 w \
x^2 &=& c_1 frac{dv}{dx} \
frac{dv}{dx} &=& c_2 x^2 \
v &=& c_3 x^3 + c_4 \
frac{y}{x} &=& c_3 x^3 + c_4 \
y &=& c_3 x^4 + c_4 x \
end{eqnarray*}
Hence the general solution is:
$$ y = C_0x^4 + C_1x $$
integration ordinary-differential-equations
asked Dec 2 '18 at 2:50
BobBob
918515
$endgroup$
1
$begingroup$
Wolfram alpha gives your answer
$endgroup$
– Kenny Lau
Dec 3 '18 at 1:28
add a comment |
$begingroup$
This problem is from the book "Introduction to Ordinary Differential Equations" by Shepley L. Ross. I am thinking I did the problem correctly because my answer matches the answer in the back of the book. What bothers me is the fact the second solution includes the first solution.
Problem:
Given that $y = x$ is a solution of
$$ x^2y'' - 4xy' + 4y = 0$$
find a linearly independent solution by reducing the order. Write the general solution.
Answer:
Let $f(x)$ represent the solution we have.
begin{eqnarray*}
f(x) &=& x \
y &=& f(x) v = xv \
y' &=& x v' + v \
y'' &=& x v'' + v' + v' = xv'' + 2v' \
end{eqnarray*}
begin{eqnarray*}
x^2(xv'' + 2v') - 4x(x v' + v) + 4xv &=& 0 \
x^3v'' + 2x^2v' - 4x^2v' &=& 0 \
x^3v'' - 2x^2v' &=& 0 \
text{Let }w &=& frac{dv}{dx} \
x^3 w' - 2x^2 w &=& 0 \
x w' - 2 w &=& 0 \
x w' &=& 2w \
x frac{dw}{dx} &=& 2w \
frac{x}{dx} &=& frac{2w}{dw} \
frac{x}{2dx} &=& frac{w}{dw} \
frac{ 2dx}{x} &=& frac{dw}{w} \
2ln{x} &=& ln{w} + c_0 \
x^2 &=& c_1 w \
x^2 &=& c_1 frac{dv}{dx} \
frac{dv}{dx} &=& c_2 x^2 \
v &=& c_3 x^3 + c_4 \
frac{y}{x} &=& c_3 x^3 + c_4 \
y &=& c_3 x^4 + c_4 x \
end{eqnarray*}
Hence the general solution is:
$$ y = C_0x^4 + C_1x $$
integration ordinary-differential-equations
asked Dec 2 '18 at 2:50
BobBob
918515
$endgroup$
This problem is from the book "Introduction to Ordinary Differential Equations" by Shepley L. Ross. I am thinking I did the problem correctly because my answer matches the answer in the back of the book. What bothers me is the fact the second solution includes the first solution.
Problem:
Given that $y = x$ is a solution of
$$ x^2y'' - 4xy' + 4y = 0$$
find a linearly independent solution by reducing the order. Write the general solution.
Answer:
Let $f(x)$ represent the solution we have.
begin{eqnarray*}
f(x) &=& x \
y &=& f(x) v = xv \
y' &=& x v' + v \
y'' &=& x v'' + v' + v' = xv'' + 2v' \
end{eqnarray*}
begin{eqnarray*}
x^2(xv'' + 2v') - 4x(x v' + v) + 4xv &=& 0 \
x^3v'' + 2x^2v' - 4x^2v' &=& 0 \
x^3v'' - 2x^2v' &=& 0 \
text{Let }w &=& frac{dv}{dx} \
x^3 w' - 2x^2 w &=& 0 \
x w' - 2 w &=& 0 \
x w' &=& 2w \
x frac{dw}{dx} &=& 2w \
frac{x}{dx} &=& frac{2w}{dw} \
frac{x}{2dx} &=& frac{w}{dw} \
frac{ 2dx}{x} &=& frac{dw}{w} \
2ln{x} &=& ln{w} + c_0 \
x^2 &=& c_1 w \
x^2 &=& c_1 frac{dv}{dx} \
frac{dv}{dx} &=& c_2 x^2 \
v &=& c_3 x^3 + c_4 \
frac{y}{x} &=& c_3 x^3 + c_4 \
y &=& c_3 x^4 + c_4 x \
end{eqnarray*}
Hence the general solution is:
$$ y = C_0x^4 + C_1x $$
integration ordinary-differential-equations
integration ordinary-differential-equations
asked Dec 2 '18 at 2:50
BobBob
918515
asked Dec 2 '18 at 2:50
BobBob
918515
asked Dec 2 '18 at 2:50
BobBob
918515
asked Dec 2 '18 at 2:50
BobBob
918515
asked Dec 2 '18 at 2:50
BobBob
918515
918515
1
$begingroup$
Wolfram alpha gives your answer
$endgroup$
– Kenny Lau
Dec 3 '18 at 1:28
add a comment |
1
$begingroup$
Wolfram alpha gives your answer
$endgroup$
– Kenny Lau
Dec 3 '18 at 1:28
1
1
$begingroup$
Wolfram alpha gives your answer
$endgroup$
– Kenny Lau
Dec 3 '18 at 1:28
$begingroup$
Wolfram alpha gives your answer
$endgroup$
– Kenny Lau
Dec 3 '18 at 1:28
add a comment |
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Wolfram alpha gives your answer
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– Kenny Lau
Dec 3 '18 at 1:28