Mixing
Solving $(ln(x)-1)y'' - frac{1}{x}y' + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x}$
$begingroup$
On my exam I had to solve the following differential equation.
begin{equation}
(ln(x)-1)y'' - frac{1}{x}y' + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}
end{equation}
Which is a differential equation of the form:
begin{equation}
y'' + a(x)y' + b(x)y = R(x)
end{equation}
The only method we've seen to solve this kind of differential equations is:
If the differential equation is of the form:
begin{equation}
y'' + a(x)y' + b(x)y = 0
end{equation}
First find a solution of the characteristic equation, being $varphi_1$. Then:
begin{equation}
varphi_2(x) = varphi_1(x)intfrac{dx}{A(x)(varphi_1(x))^2}
end{equation}
With $A(x) = e^{int a(x) dx}$
Then the homogenous solution is given by:
begin{equation}
y(x) = c_1varphi_1(x) + c_2varphi_2(x)
end{equation}
The first problem is that this doesn't satisfy the requirements for this method since the differential equation is not homogenous, but since this is the only fitting method, I'd still try to use it. My guess would be to start with the characteristic equation which gives:
begin{equation}
(ln(x)-1)x^2 - 1 + frac{1}{x^2}y = 0
end{equation}
or
begin{equation}
x^2 - frac{1}{(ln(x)-1)} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
but i wouldn't even know how to start solving this equation to find the roots of the equation. Does anyone have an idea as to how to tackle this problem.
Note the only other ways of solving linear differential equations that we have seen are ways to solve first order differential equation or ways to solve second order differential equations in the form:
begin{equation}
y'' + py' + qy = R(x);;;text{with};; p,qinmathbb{R}
end{equation}
real-analysis ordinary-differential-equations differential
asked Jan 18 at 17:53
ViktorViktor
1389
$endgroup$
add a comment |
$begingroup$
On my exam I had to solve the following differential equation.
begin{equation}
(ln(x)-1)y'' - frac{1}{x}y' + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}
end{equation}
Which is a differential equation of the form:
begin{equation}
y'' + a(x)y' + b(x)y = R(x)
end{equation}
The only method we've seen to solve this kind of differential equations is:
If the differential equation is of the form:
begin{equation}
y'' + a(x)y' + b(x)y = 0
end{equation}
First find a solution of the characteristic equation, being $varphi_1$. Then:
begin{equation}
varphi_2(x) = varphi_1(x)intfrac{dx}{A(x)(varphi_1(x))^2}
end{equation}
With $A(x) = e^{int a(x) dx}$
Then the homogenous solution is given by:
begin{equation}
y(x) = c_1varphi_1(x) + c_2varphi_2(x)
end{equation}
The first problem is that this doesn't satisfy the requirements for this method since the differential equation is not homogenous, but since this is the only fitting method, I'd still try to use it. My guess would be to start with the characteristic equation which gives:
begin{equation}
(ln(x)-1)x^2 - 1 + frac{1}{x^2}y = 0
end{equation}
or
begin{equation}
x^2 - frac{1}{(ln(x)-1)} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
but i wouldn't even know how to start solving this equation to find the roots of the equation. Does anyone have an idea as to how to tackle this problem.
Note the only other ways of solving linear differential equations that we have seen are ways to solve first order differential equation or ways to solve second order differential equations in the form:
begin{equation}
y'' + py' + qy = R(x);;;text{with};; p,qinmathbb{R}
end{equation}
real-analysis ordinary-differential-equations differential
asked Jan 18 at 17:53
ViktorViktor
1389
$endgroup$
1
$begingroup$
"The first problem is that this doesn't satisfy the requirements for this method... but since this is the only fitting method, I'd still try to use it." Why? If it doesn't satisfy the requirements, what do you hope to achieve?
$endgroup$
– nbubis
Jan 18 at 18:00
$begingroup$
Because it was a question on my exam today, so there has to be some way to solve this with the methods we have seen. Since this is the only one that resembles it (the only one with non constant coefficients) i tried to use it. I wouldn't know in what way i could possibly find a solution otherwise.
$endgroup$
– Viktor
Jan 18 at 18:01
1
$begingroup$
i think there is a typo in your equation.
$endgroup$
– Dr. Sonnhard Graubner
Jan 18 at 18:11
$begingroup$
yes it had to be x instead of x^2 in the denominator of R(x) but the rest of the equation is correct
$endgroup$
– Viktor
Jan 18 at 18:29
add a comment |
$begingroup$
On my exam I had to solve the following differential equation.
begin{equation}
(ln(x)-1)y'' - frac{1}{x}y' + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}
end{equation}
Which is a differential equation of the form:
begin{equation}
y'' + a(x)y' + b(x)y = R(x)
end{equation}
The only method we've seen to solve this kind of differential equations is:
If the differential equation is of the form:
begin{equation}
y'' + a(x)y' + b(x)y = 0
end{equation}
First find a solution of the characteristic equation, being $varphi_1$. Then:
begin{equation}
varphi_2(x) = varphi_1(x)intfrac{dx}{A(x)(varphi_1(x))^2}
end{equation}
With $A(x) = e^{int a(x) dx}$
Then the homogenous solution is given by:
begin{equation}
y(x) = c_1varphi_1(x) + c_2varphi_2(x)
end{equation}
The first problem is that this doesn't satisfy the requirements for this method since the differential equation is not homogenous, but since this is the only fitting method, I'd still try to use it. My guess would be to start with the characteristic equation which gives:
begin{equation}
(ln(x)-1)x^2 - 1 + frac{1}{x^2}y = 0
end{equation}
or
begin{equation}
x^2 - frac{1}{(ln(x)-1)} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
but i wouldn't even know how to start solving this equation to find the roots of the equation. Does anyone have an idea as to how to tackle this problem.
Note the only other ways of solving linear differential equations that we have seen are ways to solve first order differential equation or ways to solve second order differential equations in the form:
begin{equation}
y'' + py' + qy = R(x);;;text{with};; p,qinmathbb{R}
end{equation}
real-analysis ordinary-differential-equations differential
asked Jan 18 at 17:53
ViktorViktor
1389
$endgroup$
On my exam I had to solve the following differential equation.
begin{equation}
(ln(x)-1)y'' - frac{1}{x}y' + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}
end{equation}
Which is a differential equation of the form:
begin{equation}
y'' + a(x)y' + b(x)y = R(x)
end{equation}
The only method we've seen to solve this kind of differential equations is:
If the differential equation is of the form:
begin{equation}
y'' + a(x)y' + b(x)y = 0
end{equation}
First find a solution of the characteristic equation, being $varphi_1$. Then:
begin{equation}
varphi_2(x) = varphi_1(x)intfrac{dx}{A(x)(varphi_1(x))^2}
end{equation}
With $A(x) = e^{int a(x) dx}$
Then the homogenous solution is given by:
begin{equation}
y(x) = c_1varphi_1(x) + c_2varphi_2(x)
end{equation}
The first problem is that this doesn't satisfy the requirements for this method since the differential equation is not homogenous, but since this is the only fitting method, I'd still try to use it. My guess would be to start with the characteristic equation which gives:
begin{equation}
(ln(x)-1)x^2 - 1 + frac{1}{x^2}y = 0
end{equation}
or
begin{equation}
x^2 - frac{1}{(ln(x)-1)} + frac{1}{x^2(ln(x)-1)} = 0
end{equation}
but i wouldn't even know how to start solving this equation to find the roots of the equation. Does anyone have an idea as to how to tackle this problem.
Note the only other ways of solving linear differential equations that we have seen are ways to solve first order differential equation or ways to solve second order differential equations in the form:
begin{equation}
y'' + py' + qy = R(x);;;text{with};; p,qinmathbb{R}
end{equation}
real-analysis ordinary-differential-equations differential
real-analysis ordinary-differential-equations differential
asked Jan 18 at 17:53
ViktorViktor
1389
asked Jan 18 at 17:53
ViktorViktor
1389
edited Jan 18 at 18:28
Viktor
asked Jan 18 at 17:53
ViktorViktor
1389
asked Jan 18 at 17:53
ViktorViktor
1389
asked Jan 18 at 17:53
ViktorViktor
1389
1389
1
$begingroup$
"The first problem is that this doesn't satisfy the requirements for this method... but since this is the only fitting method, I'd still try to use it." Why? If it doesn't satisfy the requirements, what do you hope to achieve?
$endgroup$
– nbubis
Jan 18 at 18:00
$begingroup$
Because it was a question on my exam today, so there has to be some way to solve this with the methods we have seen. Since this is the only one that resembles it (the only one with non constant coefficients) i tried to use it. I wouldn't know in what way i could possibly find a solution otherwise.
$endgroup$
– Viktor
Jan 18 at 18:01
1
$begingroup$
i think there is a typo in your equation.
$endgroup$
– Dr. Sonnhard Graubner
Jan 18 at 18:11
$begingroup$
yes it had to be x instead of x^2 in the denominator of R(x) but the rest of the equation is correct
$endgroup$
– Viktor
Jan 18 at 18:29
add a comment |
1
$begingroup$
"The first problem is that this doesn't satisfy the requirements for this method... but since this is the only fitting method, I'd still try to use it." Why? If it doesn't satisfy the requirements, what do you hope to achieve?
$endgroup$
– nbubis
Jan 18 at 18:00
$begingroup$
Because it was a question on my exam today, so there has to be some way to solve this with the methods we have seen. Since this is the only one that resembles it (the only one with non constant coefficients) i tried to use it. I wouldn't know in what way i could possibly find a solution otherwise.
$endgroup$
– Viktor
Jan 18 at 18:01
1
$begingroup$
i think there is a typo in your equation.
$endgroup$
– Dr. Sonnhard Graubner
Jan 18 at 18:11
$begingroup$
yes it had to be x instead of x^2 in the denominator of R(x) but the rest of the equation is correct
$endgroup$
– Viktor
Jan 18 at 18:29
1
1
$begingroup$
"The first problem is that this doesn't satisfy the requirements for this method... but since this is the only fitting method, I'd still try to use it." Why? If it doesn't satisfy the requirements, what do you hope to achieve?
$endgroup$
– nbubis
Jan 18 at 18:00
$begingroup$
"The first problem is that this doesn't satisfy the requirements for this method... but since this is the only fitting method, I'd still try to use it." Why? If it doesn't satisfy the requirements, what do you hope to achieve?
$endgroup$
– nbubis
Jan 18 at 18:00
$begingroup$
Because it was a question on my exam today, so there has to be some way to solve this with the methods we have seen. Since this is the only one that resembles it (the only one with non constant coefficients) i tried to use it. I wouldn't know in what way i could possibly find a solution otherwise.
$endgroup$
– Viktor
Jan 18 at 18:01
$begingroup$
Because it was a question on my exam today, so there has to be some way to solve this with the methods we have seen. Since this is the only one that resembles it (the only one with non constant coefficients) i tried to use it. I wouldn't know in what way i could possibly find a solution otherwise.
$endgroup$
– Viktor
Jan 18 at 18:01
1
1
$begingroup$
i think there is a typo in your equation.
$endgroup$
– Dr. Sonnhard Graubner
Jan 18 at 18:11
$begingroup$
i think there is a typo in your equation.
$endgroup$
– Dr. Sonnhard Graubner
Jan 18 at 18:11
$begingroup$
yes it had to be x instead of x^2 in the denominator of R(x) but the rest of the equation is correct
$endgroup$
– Viktor
Jan 18 at 18:29
$begingroup$
yes it had to be x instead of x^2 in the denominator of R(x) but the rest of the equation is correct
$endgroup$
– Viktor
Jan 18 at 18:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Replace
$$begin{align}z&=ln x -1\
w(z)&=y(x)text{.}end{align}$$
Then the differential equation is
$$z w'' -(z+1)w'+w=z^2mathrm{e}^{z+1}text{.}$$
The linear differential operator on the left side factors, so the solution to this equation can be found by solving in turn the first-order equations
$$begin{align}
zv'-v&=z^2mathrm{e}^{z+1} \
w'-w&=vtext{.}
end{align}$$
Another form for these equations is
$$begin{align}
left(tfrac{v}{z}right)'&=mathrm{e}^{z+1} \
(mathrm{e}^{-z}w)'&=mathrm{e}^{-z}vtext{.}
end{align}$$
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
$endgroup$
add a comment |
$begingroup$
Hint: You can find by trial-and-error that the two homogeneous solutions are
$ varphi_1(x) = ln x $ and $varphi_2(x) = x$
Next, use variation of parameters to find a particular solution of the form
$$ y(x) = u_1(x)varphi_1(x) + u_2(x)varphi_2(x) $$
Then you have a system of equations
begin{align}
u_1'(x)varphi_1(x) + u_2'(x)varphi_2(x) &= 0 \
u_1'(x)varphi_1'(x) + u_2'(x)varphi_2'(x) &= R(x)
end{align}
where $R(x) = frac{ln x -1}{x^2}$
Solve thiss for $u_1'(x)$ and $u_2'(x)$, then integrate.
answered Jan 19 at 10:31
DylanDylan
13.4k31027
$endgroup$
add a comment |
$begingroup$
$$(ln(x)-1)frac{d^2y}{dx^2} - frac{1}{x}frac{dy}{dx} + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}$$
Change of variable : $t=ln(x)-1quad;quad x=e^{t+1}quad;quad dx=x:dt$
$frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=frac{1}{x}frac{dy}{dt}$
$frac{d^2y}{dx^2}=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x}frac{d^2y}{dt^2}frac{dt}{dx}= -frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}$
$$t(-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}) -frac{1}{x}(frac{1}{x}frac{dy}{dt})+ frac{1}{x^2}y = frac{t^2}{x^2}$$
$$tfrac{d^2y}{dt^2}-(t+1)frac{dy}{dt}+y=t^2 $$
There is no difficulty to solve this second order linear ODE. Obvious solutions of the associated homogeneous ODE are $(t+1)$ and $e^t$. A particular solution of the ODE is $-t^2$ . This leads to :
$$y= c_1e^t+c_2(t+1)-t^2$$
Finally :
$y(x)= c_1e^{ln(x)-1}+c_2(ln(x)-1+1)-(ln(x)-1)^2$
$$y(x)= c_1e^{-1}x+c_2ln(x)-(ln(x)-1)^2$$
or on equivalent form :
$$y(x)= C_1x+C_2ln(x)-(ln(x))^2-1$$
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078569%2fsolving-lnx-1y-frac1xy-frac1x2y-frac-lnx-12%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Replace
$$begin{align}z&=ln x -1\
w(z)&=y(x)text{.}end{align}$$
Then the differential equation is
$$z w'' -(z+1)w'+w=z^2mathrm{e}^{z+1}text{.}$$
The linear differential operator on the left side factors, so the solution to this equation can be found by solving in turn the first-order equations
$$begin{align}
zv'-v&=z^2mathrm{e}^{z+1} \
w'-w&=vtext{.}
end{align}$$
Another form for these equations is
$$begin{align}
left(tfrac{v}{z}right)'&=mathrm{e}^{z+1} \
(mathrm{e}^{-z}w)'&=mathrm{e}^{-z}vtext{.}
end{align}$$
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
$endgroup$
add a comment |
$begingroup$
Replace
$$begin{align}z&=ln x -1\
w(z)&=y(x)text{.}end{align}$$
Then the differential equation is
$$z w'' -(z+1)w'+w=z^2mathrm{e}^{z+1}text{.}$$
The linear differential operator on the left side factors, so the solution to this equation can be found by solving in turn the first-order equations
$$begin{align}
zv'-v&=z^2mathrm{e}^{z+1} \
w'-w&=vtext{.}
end{align}$$
Another form for these equations is
$$begin{align}
left(tfrac{v}{z}right)'&=mathrm{e}^{z+1} \
(mathrm{e}^{-z}w)'&=mathrm{e}^{-z}vtext{.}
end{align}$$
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
$endgroup$
add a comment |
$begingroup$
Replace
$$begin{align}z&=ln x -1\
w(z)&=y(x)text{.}end{align}$$
Then the differential equation is
$$z w'' -(z+1)w'+w=z^2mathrm{e}^{z+1}text{.}$$
The linear differential operator on the left side factors, so the solution to this equation can be found by solving in turn the first-order equations
$$begin{align}
zv'-v&=z^2mathrm{e}^{z+1} \
w'-w&=vtext{.}
end{align}$$
Another form for these equations is
$$begin{align}
left(tfrac{v}{z}right)'&=mathrm{e}^{z+1} \
(mathrm{e}^{-z}w)'&=mathrm{e}^{-z}vtext{.}
end{align}$$
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
$endgroup$
Replace
$$begin{align}z&=ln x -1\
w(z)&=y(x)text{.}end{align}$$
Then the differential equation is
$$z w'' -(z+1)w'+w=z^2mathrm{e}^{z+1}text{.}$$
The linear differential operator on the left side factors, so the solution to this equation can be found by solving in turn the first-order equations
$$begin{align}
zv'-v&=z^2mathrm{e}^{z+1} \
w'-w&=vtext{.}
end{align}$$
Another form for these equations is
$$begin{align}
left(tfrac{v}{z}right)'&=mathrm{e}^{z+1} \
(mathrm{e}^{-z}w)'&=mathrm{e}^{-z}vtext{.}
end{align}$$
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
edited Jan 18 at 18:36
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
answered Jan 18 at 18:30
K B DaveK B Dave
3,492317
3,492317
add a comment |
add a comment |
$begingroup$
Hint: You can find by trial-and-error that the two homogeneous solutions are
$ varphi_1(x) = ln x $ and $varphi_2(x) = x$
Next, use variation of parameters to find a particular solution of the form
$$ y(x) = u_1(x)varphi_1(x) + u_2(x)varphi_2(x) $$
Then you have a system of equations
begin{align}
u_1'(x)varphi_1(x) + u_2'(x)varphi_2(x) &= 0 \
u_1'(x)varphi_1'(x) + u_2'(x)varphi_2'(x) &= R(x)
end{align}
where $R(x) = frac{ln x -1}{x^2}$
Solve thiss for $u_1'(x)$ and $u_2'(x)$, then integrate.
answered Jan 19 at 10:31
DylanDylan
13.4k31027
$endgroup$
add a comment |
$begingroup$
Hint: You can find by trial-and-error that the two homogeneous solutions are
$ varphi_1(x) = ln x $ and $varphi_2(x) = x$
Next, use variation of parameters to find a particular solution of the form
$$ y(x) = u_1(x)varphi_1(x) + u_2(x)varphi_2(x) $$
Then you have a system of equations
begin{align}
u_1'(x)varphi_1(x) + u_2'(x)varphi_2(x) &= 0 \
u_1'(x)varphi_1'(x) + u_2'(x)varphi_2'(x) &= R(x)
end{align}
where $R(x) = frac{ln x -1}{x^2}$
Solve thiss for $u_1'(x)$ and $u_2'(x)$, then integrate.
answered Jan 19 at 10:31
DylanDylan
13.4k31027
$endgroup$
add a comment |
$begingroup$
Hint: You can find by trial-and-error that the two homogeneous solutions are
$ varphi_1(x) = ln x $ and $varphi_2(x) = x$
Next, use variation of parameters to find a particular solution of the form
$$ y(x) = u_1(x)varphi_1(x) + u_2(x)varphi_2(x) $$
Then you have a system of equations
begin{align}
u_1'(x)varphi_1(x) + u_2'(x)varphi_2(x) &= 0 \
u_1'(x)varphi_1'(x) + u_2'(x)varphi_2'(x) &= R(x)
end{align}
where $R(x) = frac{ln x -1}{x^2}$
Solve thiss for $u_1'(x)$ and $u_2'(x)$, then integrate.
answered Jan 19 at 10:31
DylanDylan
13.4k31027
$endgroup$
Hint: You can find by trial-and-error that the two homogeneous solutions are
$ varphi_1(x) = ln x $ and $varphi_2(x) = x$
Next, use variation of parameters to find a particular solution of the form
$$ y(x) = u_1(x)varphi_1(x) + u_2(x)varphi_2(x) $$
Then you have a system of equations
begin{align}
u_1'(x)varphi_1(x) + u_2'(x)varphi_2(x) &= 0 \
u_1'(x)varphi_1'(x) + u_2'(x)varphi_2'(x) &= R(x)
end{align}
where $R(x) = frac{ln x -1}{x^2}$
Solve thiss for $u_1'(x)$ and $u_2'(x)$, then integrate.
answered Jan 19 at 10:31
DylanDylan
13.4k31027
answered Jan 19 at 10:31
DylanDylan
13.4k31027
answered Jan 19 at 10:31
DylanDylan
13.4k31027
answered Jan 19 at 10:31
DylanDylan
13.4k31027
13.4k31027
add a comment |
add a comment |
$begingroup$
$$(ln(x)-1)frac{d^2y}{dx^2} - frac{1}{x}frac{dy}{dx} + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}$$
Change of variable : $t=ln(x)-1quad;quad x=e^{t+1}quad;quad dx=x:dt$
$frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=frac{1}{x}frac{dy}{dt}$
$frac{d^2y}{dx^2}=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x}frac{d^2y}{dt^2}frac{dt}{dx}= -frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}$
$$t(-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}) -frac{1}{x}(frac{1}{x}frac{dy}{dt})+ frac{1}{x^2}y = frac{t^2}{x^2}$$
$$tfrac{d^2y}{dt^2}-(t+1)frac{dy}{dt}+y=t^2 $$
There is no difficulty to solve this second order linear ODE. Obvious solutions of the associated homogeneous ODE are $(t+1)$ and $e^t$. A particular solution of the ODE is $-t^2$ . This leads to :
$$y= c_1e^t+c_2(t+1)-t^2$$
Finally :
$y(x)= c_1e^{ln(x)-1}+c_2(ln(x)-1+1)-(ln(x)-1)^2$
$$y(x)= c_1e^{-1}x+c_2ln(x)-(ln(x)-1)^2$$
or on equivalent form :
$$y(x)= C_1x+C_2ln(x)-(ln(x))^2-1$$
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
$endgroup$
add a comment |
$begingroup$
$$(ln(x)-1)frac{d^2y}{dx^2} - frac{1}{x}frac{dy}{dx} + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}$$
Change of variable : $t=ln(x)-1quad;quad x=e^{t+1}quad;quad dx=x:dt$
$frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=frac{1}{x}frac{dy}{dt}$
$frac{d^2y}{dx^2}=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x}frac{d^2y}{dt^2}frac{dt}{dx}= -frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}$
$$t(-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}) -frac{1}{x}(frac{1}{x}frac{dy}{dt})+ frac{1}{x^2}y = frac{t^2}{x^2}$$
$$tfrac{d^2y}{dt^2}-(t+1)frac{dy}{dt}+y=t^2 $$
There is no difficulty to solve this second order linear ODE. Obvious solutions of the associated homogeneous ODE are $(t+1)$ and $e^t$. A particular solution of the ODE is $-t^2$ . This leads to :
$$y= c_1e^t+c_2(t+1)-t^2$$
Finally :
$y(x)= c_1e^{ln(x)-1}+c_2(ln(x)-1+1)-(ln(x)-1)^2$
$$y(x)= c_1e^{-1}x+c_2ln(x)-(ln(x)-1)^2$$
or on equivalent form :
$$y(x)= C_1x+C_2ln(x)-(ln(x))^2-1$$
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
$endgroup$
add a comment |
$begingroup$
$$(ln(x)-1)frac{d^2y}{dx^2} - frac{1}{x}frac{dy}{dx} + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}$$
Change of variable : $t=ln(x)-1quad;quad x=e^{t+1}quad;quad dx=x:dt$
$frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=frac{1}{x}frac{dy}{dt}$
$frac{d^2y}{dx^2}=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x}frac{d^2y}{dt^2}frac{dt}{dx}= -frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}$
$$t(-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}) -frac{1}{x}(frac{1}{x}frac{dy}{dt})+ frac{1}{x^2}y = frac{t^2}{x^2}$$
$$tfrac{d^2y}{dt^2}-(t+1)frac{dy}{dt}+y=t^2 $$
There is no difficulty to solve this second order linear ODE. Obvious solutions of the associated homogeneous ODE are $(t+1)$ and $e^t$. A particular solution of the ODE is $-t^2$ . This leads to :
$$y= c_1e^t+c_2(t+1)-t^2$$
Finally :
$y(x)= c_1e^{ln(x)-1}+c_2(ln(x)-1+1)-(ln(x)-1)^2$
$$y(x)= c_1e^{-1}x+c_2ln(x)-(ln(x)-1)^2$$
or on equivalent form :
$$y(x)= C_1x+C_2ln(x)-(ln(x))^2-1$$
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
$endgroup$
$$(ln(x)-1)frac{d^2y}{dx^2} - frac{1}{x}frac{dy}{dx} + frac{1}{x^2}y = frac{(ln(x) - 1)^2}{x^2}$$
Change of variable : $t=ln(x)-1quad;quad x=e^{t+1}quad;quad dx=x:dt$
$frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=frac{1}{x}frac{dy}{dt}$
$frac{d^2y}{dx^2}=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x}frac{d^2y}{dt^2}frac{dt}{dx}= -frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}$
$$t(-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}) -frac{1}{x}(frac{1}{x}frac{dy}{dt})+ frac{1}{x^2}y = frac{t^2}{x^2}$$
$$tfrac{d^2y}{dt^2}-(t+1)frac{dy}{dt}+y=t^2 $$
There is no difficulty to solve this second order linear ODE. Obvious solutions of the associated homogeneous ODE are $(t+1)$ and $e^t$. A particular solution of the ODE is $-t^2$ . This leads to :
$$y= c_1e^t+c_2(t+1)-t^2$$
Finally :
$y(x)= c_1e^{ln(x)-1}+c_2(ln(x)-1+1)-(ln(x)-1)^2$
$$y(x)= c_1e^{-1}x+c_2ln(x)-(ln(x)-1)^2$$
or on equivalent form :
$$y(x)= C_1x+C_2ln(x)-(ln(x))^2-1$$
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
edited Jan 19 at 11:33
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
answered Jan 19 at 11:22
JJacquelinJJacquelin
44.2k21853
44.2k21853
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078569%2fsolving-lnx-1y-frac1xy-frac1x2y-frac-lnx-12%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
"The first problem is that this doesn't satisfy the requirements for this method... but since this is the only fitting method, I'd still try to use it." Why? If it doesn't satisfy the requirements, what do you hope to achieve?
$endgroup$
– nbubis
Jan 18 at 18:00
$begingroup$
Because it was a question on my exam today, so there has to be some way to solve this with the methods we have seen. Since this is the only one that resembles it (the only one with non constant coefficients) i tried to use it. I wouldn't know in what way i could possibly find a solution otherwise.
$endgroup$
– Viktor
Jan 18 at 18:01
1
$begingroup$
i think there is a typo in your equation.
$endgroup$
– Dr. Sonnhard Graubner
Jan 18 at 18:11
$begingroup$
yes it had to be x instead of x^2 in the denominator of R(x) but the rest of the equation is correct
$endgroup$
– Viktor
Jan 18 at 18:29