general solution of $xfrac{dy}{dx}-4y=x^6e^x$
$begingroup$
I have a question about the following example in Zill and Wright's textbook:
Example $quad$ Solve $xfrac{dy}{dx}-4y=x^6e^x.$
This is a simple example to illustrate how to find general solution of a linear first-order differential equation (DE):
Solution $quad$ By dividing by $x$ we get the standard form
$$frac{dy}{dx}-frac{4}{x}y=x^5e^x.$$
Let $P(x)triangleq-4/x$ and $f(x)triangleq x^5e^x$. Note that $P$ and $f$ are continuous on $(0, infty)$, so the general solution on $(0, infty)$ is $y=y_c+y_p,$ where $y_c=ce^{-int P(x)dx}$ is the general solution to the homogeneneous DE, and $$y_p=e^{-int P(x)dx}int e^{int P(x)dx}f(x)dx$$
is a particular solution to the non-homogeneous DE.
Now, note that $e^{-int P(x)dx}=e^{int 4/xdx}=x^{4}$ on $(0, infty)$. So $y_c=cx^4$ and a particular solution is $y_p=x^{4}int xe^xdx=x^4(x-1)e^x.$ Hence the general solution is $y=cx^4+x^4(x-1)e^x$ on $(0, infty)$.
My question: Isn't $y=cx^4+x^4(x-1)e^x$ the general solution on the entire real line as well?
We're restricted to $(0, infty)$ to make $P$ continuous. The above argument also holds on $(-infty, 0)$. However, $P$ isn't in the original problem. It's an intermediate term we invented to solve the DE, and as a result we exclude ${0}$ from the real line.
Looking at the original DE, any solution $y$ must be $0$ when $x=0.$ And $y=cx^4+x^4(x-1)e^x$ clearly satisfies this. So is there any problem to say that it is the general solution of the original DE on the entire real line? Did I miss something? Thanks a lot!
ordinary-differential-equations proof-verification
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
$endgroup$
add a comment |
$begingroup$
I have a question about the following example in Zill and Wright's textbook:
Example $quad$ Solve $xfrac{dy}{dx}-4y=x^6e^x.$
This is a simple example to illustrate how to find general solution of a linear first-order differential equation (DE):
Solution $quad$ By dividing by $x$ we get the standard form
$$frac{dy}{dx}-frac{4}{x}y=x^5e^x.$$
Let $P(x)triangleq-4/x$ and $f(x)triangleq x^5e^x$. Note that $P$ and $f$ are continuous on $(0, infty)$, so the general solution on $(0, infty)$ is $y=y_c+y_p,$ where $y_c=ce^{-int P(x)dx}$ is the general solution to the homogeneneous DE, and $$y_p=e^{-int P(x)dx}int e^{int P(x)dx}f(x)dx$$
is a particular solution to the non-homogeneous DE.
Now, note that $e^{-int P(x)dx}=e^{int 4/xdx}=x^{4}$ on $(0, infty)$. So $y_c=cx^4$ and a particular solution is $y_p=x^{4}int xe^xdx=x^4(x-1)e^x.$ Hence the general solution is $y=cx^4+x^4(x-1)e^x$ on $(0, infty)$.
My question: Isn't $y=cx^4+x^4(x-1)e^x$ the general solution on the entire real line as well?
We're restricted to $(0, infty)$ to make $P$ continuous. The above argument also holds on $(-infty, 0)$. However, $P$ isn't in the original problem. It's an intermediate term we invented to solve the DE, and as a result we exclude ${0}$ from the real line.
Looking at the original DE, any solution $y$ must be $0$ when $x=0.$ And $y=cx^4+x^4(x-1)e^x$ clearly satisfies this. So is there any problem to say that it is the general solution of the original DE on the entire real line? Did I miss something? Thanks a lot!
ordinary-differential-equations proof-verification
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
$endgroup$
add a comment |
$begingroup$
I have a question about the following example in Zill and Wright's textbook:
Example $quad$ Solve $xfrac{dy}{dx}-4y=x^6e^x.$
This is a simple example to illustrate how to find general solution of a linear first-order differential equation (DE):
Solution $quad$ By dividing by $x$ we get the standard form
$$frac{dy}{dx}-frac{4}{x}y=x^5e^x.$$
Let $P(x)triangleq-4/x$ and $f(x)triangleq x^5e^x$. Note that $P$ and $f$ are continuous on $(0, infty)$, so the general solution on $(0, infty)$ is $y=y_c+y_p,$ where $y_c=ce^{-int P(x)dx}$ is the general solution to the homogeneneous DE, and $$y_p=e^{-int P(x)dx}int e^{int P(x)dx}f(x)dx$$
is a particular solution to the non-homogeneous DE.
Now, note that $e^{-int P(x)dx}=e^{int 4/xdx}=x^{4}$ on $(0, infty)$. So $y_c=cx^4$ and a particular solution is $y_p=x^{4}int xe^xdx=x^4(x-1)e^x.$ Hence the general solution is $y=cx^4+x^4(x-1)e^x$ on $(0, infty)$.
My question: Isn't $y=cx^4+x^4(x-1)e^x$ the general solution on the entire real line as well?
We're restricted to $(0, infty)$ to make $P$ continuous. The above argument also holds on $(-infty, 0)$. However, $P$ isn't in the original problem. It's an intermediate term we invented to solve the DE, and as a result we exclude ${0}$ from the real line.
Looking at the original DE, any solution $y$ must be $0$ when $x=0.$ And $y=cx^4+x^4(x-1)e^x$ clearly satisfies this. So is there any problem to say that it is the general solution of the original DE on the entire real line? Did I miss something? Thanks a lot!
ordinary-differential-equations proof-verification
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
$endgroup$
I have a question about the following example in Zill and Wright's textbook:
Example $quad$ Solve $xfrac{dy}{dx}-4y=x^6e^x.$
This is a simple example to illustrate how to find general solution of a linear first-order differential equation (DE):
Solution $quad$ By dividing by $x$ we get the standard form
$$frac{dy}{dx}-frac{4}{x}y=x^5e^x.$$
Let $P(x)triangleq-4/x$ and $f(x)triangleq x^5e^x$. Note that $P$ and $f$ are continuous on $(0, infty)$, so the general solution on $(0, infty)$ is $y=y_c+y_p,$ where $y_c=ce^{-int P(x)dx}$ is the general solution to the homogeneneous DE, and $$y_p=e^{-int P(x)dx}int e^{int P(x)dx}f(x)dx$$
is a particular solution to the non-homogeneous DE.
Now, note that $e^{-int P(x)dx}=e^{int 4/xdx}=x^{4}$ on $(0, infty)$. So $y_c=cx^4$ and a particular solution is $y_p=x^{4}int xe^xdx=x^4(x-1)e^x.$ Hence the general solution is $y=cx^4+x^4(x-1)e^x$ on $(0, infty)$.
My question: Isn't $y=cx^4+x^4(x-1)e^x$ the general solution on the entire real line as well?
We're restricted to $(0, infty)$ to make $P$ continuous. The above argument also holds on $(-infty, 0)$. However, $P$ isn't in the original problem. It's an intermediate term we invented to solve the DE, and as a result we exclude ${0}$ from the real line.
Looking at the original DE, any solution $y$ must be $0$ when $x=0.$ And $y=cx^4+x^4(x-1)e^x$ clearly satisfies this. So is there any problem to say that it is the general solution of the original DE on the entire real line? Did I miss something? Thanks a lot!
ordinary-differential-equations proof-verification
ordinary-differential-equations proof-verification
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
asked Dec 6 '18 at 4:20
syeh_106syeh_106
1,137813
1,137813
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Because of the singular point at 0, the solution on each side doesn't necessarily have to "match up". That means $c$ can take on different values for positive and negative numbers, and the general solution is
$$
y = begin{cases}c_1 x^4 + x^4(x-1)e^x & x < 0 \ c_2x^4 + x^4(x-1)e^x & xge 0end{cases}
$$
or equivalently
$$
y = b_1 x^4 + b_2 x^3|x| + x^4(x-1)e^x
$$
for some constants $b_1$ and $b_2$.
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
$endgroup$
1
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
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%2f3028027%2fgeneral-solution-of-x-fracdydx-4y-x6ex%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because of the singular point at 0, the solution on each side doesn't necessarily have to "match up". That means $c$ can take on different values for positive and negative numbers, and the general solution is
$$
y = begin{cases}c_1 x^4 + x^4(x-1)e^x & x < 0 \ c_2x^4 + x^4(x-1)e^x & xge 0end{cases}
$$
or equivalently
$$
y = b_1 x^4 + b_2 x^3|x| + x^4(x-1)e^x
$$
for some constants $b_1$ and $b_2$.
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
$endgroup$
1
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
add a comment |
$begingroup$
Because of the singular point at 0, the solution on each side doesn't necessarily have to "match up". That means $c$ can take on different values for positive and negative numbers, and the general solution is
$$
y = begin{cases}c_1 x^4 + x^4(x-1)e^x & x < 0 \ c_2x^4 + x^4(x-1)e^x & xge 0end{cases}
$$
or equivalently
$$
y = b_1 x^4 + b_2 x^3|x| + x^4(x-1)e^x
$$
for some constants $b_1$ and $b_2$.
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
$endgroup$
1
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
add a comment |
$begingroup$
Because of the singular point at 0, the solution on each side doesn't necessarily have to "match up". That means $c$ can take on different values for positive and negative numbers, and the general solution is
$$
y = begin{cases}c_1 x^4 + x^4(x-1)e^x & x < 0 \ c_2x^4 + x^4(x-1)e^x & xge 0end{cases}
$$
or equivalently
$$
y = b_1 x^4 + b_2 x^3|x| + x^4(x-1)e^x
$$
for some constants $b_1$ and $b_2$.
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
$endgroup$
Because of the singular point at 0, the solution on each side doesn't necessarily have to "match up". That means $c$ can take on different values for positive and negative numbers, and the general solution is
$$
y = begin{cases}c_1 x^4 + x^4(x-1)e^x & x < 0 \ c_2x^4 + x^4(x-1)e^x & xge 0end{cases}
$$
or equivalently
$$
y = b_1 x^4 + b_2 x^3|x| + x^4(x-1)e^x
$$
for some constants $b_1$ and $b_2$.
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
answered Dec 6 '18 at 4:31
eyeballfrogeyeballfrog
6,103629
6,103629
1
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
add a comment |
1
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
1
1
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
A singular point is any point where one of the coefficient functions of the DE in standard form has a singularity. In this case, $4/x$ has a singularity at $0$.
$endgroup$
– eyeballfrog
Dec 6 '18 at 4:56
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
You mean to say original DE is not continuous at x=0. Am i right?
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 5:44
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
$begingroup$
Indeed. I thought about this question some more. So, for instance, the initial value problem (IVP), $x frac{dy}{dx}-4y=x^6e^x, y(1)=1$, has a unique solution only on $(0, infty)$ or $[0, infty)$, but not on the entire real line? In other words, $c_2=1$ is uniquely defined, but $c_1$ can take any value. For example, together with either $c_1=3$ or $c_1=-2$, we have a legitimate solution on $(-infty, infty)$ to the IVP. And each solution is continuously differentiable on $(-infty, infty)$.
$endgroup$
– syeh_106
Dec 7 '18 at 2:46
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%2f3028027%2fgeneral-solution-of-x-fracdydx-4y-x6ex%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
