Mixing
Solution to Differential equation 2nd order, x*e^x
$begingroup$
I'm trying to solve the following differential equation:
$y'' + y = x cdot e^x$
I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...
Can somebody explain to me how to solve this?
Thanks.
Edit: Here's the solution:
$y_p = (a cdot x + b) cdot e^x$
$y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$
$y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$
$Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$
$Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$
All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$
ordinary-differential-equations exponential-function
asked Jan 21 at 15:09
zutruzutru
213
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the following differential equation:
$y'' + y = x cdot e^x$
I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...
Can somebody explain to me how to solve this?
Thanks.
Edit: Here's the solution:
$y_p = (a cdot x + b) cdot e^x$
$y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$
$y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$
$Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$
$Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$
All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$
ordinary-differential-equations exponential-function
asked Jan 21 at 15:09
zutruzutru
213
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the following differential equation:
$y'' + y = x cdot e^x$
I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...
Can somebody explain to me how to solve this?
Thanks.
Edit: Here's the solution:
$y_p = (a cdot x + b) cdot e^x$
$y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$
$y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$
$Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$
$Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$
All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$
ordinary-differential-equations exponential-function
asked Jan 21 at 15:09
zutruzutru
213
$endgroup$
I'm trying to solve the following differential equation:
$y'' + y = x cdot e^x$
I already have the homogeneous solution: $y_h = c_1 cdot cos(x) + c_2 cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a cdot x cdot e^x$ but that didn't quite work, or maybe I did something wrong...
Can somebody explain to me how to solve this?
Thanks.
Edit: Here's the solution:
$y_p = (a cdot x + b) cdot e^x$
$y_p' = (a + a cdot x + b) cdot e^x$ and $ y_p'' = (2 cdot a + a cdot x + b) cdot e^x$
$y_p'' + y_p = (2 cdot a cdot x + 2 cdot a + 2 cdot b) cdot e^x$
$Rightarrow 2 cdot a = 1 Rightarrow a = frac{1}{2}$ and $2 cdot a + 2 cdot b = 0 Leftrightarrow 2 cdot a = 1 = - 2 cdot b Rightarrow b = - frac{1}{2}$
$Rightarrow y_p = (a cdot x + b) cdot e^x = frac{x-1}{2} cdot e^x$
All in all: $y = y_h + y_p = c_1 cdot cos(x) + c_2 cdot sin(x) + frac{x-1}{2} cdot e^x$
ordinary-differential-equations exponential-function
ordinary-differential-equations exponential-function
asked Jan 21 at 15:09
zutruzutru
213
asked Jan 21 at 15:09
zutruzutru
213
edited Jan 21 at 17:11
zutru
asked Jan 21 at 15:09
zutruzutru
213
asked Jan 21 at 15:09
zutruzutru
213
asked Jan 21 at 15:09
zutruzutru
213
213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
answered Jan 21 at 15:12
KlausKlaus
2,12711
$endgroup$
1
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
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%2f3081976%2fsolution-to-differential-equation-2nd-order-xex%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
$begingroup$
Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
answered Jan 21 at 15:12
KlausKlaus
2,12711
$endgroup$
1
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
add a comment |
$begingroup$
Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
answered Jan 21 at 15:12
KlausKlaus
2,12711
$endgroup$
1
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
add a comment |
$begingroup$
Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
answered Jan 21 at 15:12
KlausKlaus
2,12711
$endgroup$
Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
answered Jan 21 at 15:12
KlausKlaus
2,12711
answered Jan 21 at 15:12
KlausKlaus
2,12711
answered Jan 21 at 15:12
KlausKlaus
2,12711
answered Jan 21 at 15:12
KlausKlaus
2,12711
2,12711
1
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
add a comment |
1
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
1
1
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
$begingroup$
oh.. yeah, that works, thanks
$endgroup$
– zutru
Jan 21 at 17:11
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%2f3081976%2fsolution-to-differential-equation-2nd-order-xex%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
