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1. Shard Calculation

Query:

Response:

Calculated Shard: 198 (from laksa055)

2. Crawled Status Check

Query:

curl -X POST \
  'http://laksa198.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://math-physics-problems.fandom.com/wiki/Laplace_Transform\')), getAhrefsUnparsedNoserviceFromURL(\'https://math-physics-problems.fandom.com/wiki/Laplace_Transform\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/math-physics-problems.fandom.com\/wiki\/Laplace_Transform","crawl_time":1768546638,"first_indexed_time":0,"http_code":200,"src_unparsed":"com,fandom!math-physics-problems,\/wiki\/Laplace_Transform s443","src_root_hash":"11439660390604180198","history_drop_reason":null,"meta_title":"Laplace Transform | Math & Physics Problems Wikia | Fandom","meta_descriptions":["Solve the differential equation y ″ − y ′ − 2 y = 4 e − t {\\displaystyle y'' - y'- 2y = 4{e}^{-t} } subject to the initial-values y ( 0 ) = 0 {\\displaystyle y(0) = 0 } and y ′ ( 0 ) = 0. {\\displaystyle y'(0)=0.} Check the table of Laplace transforms (Figure 1) for the relevant Laplace..."],"attrs_boilerpipe_text":"Problem\n[\n]\nSolve the differential equation\nsubject  to the initial-values\nand\nFigure 1. Illustrative description of the Laplace transform\nSolution\n[\n]\nFigure 2. Table of Laplace transforms\nCheck the table of Laplace transforms (Figure 1) for the relevant Laplace transformations. The following Laplace transforms will be useful for this differential equation.\nThis converts the differential equation into the following equation\nSolving for\ngives\nBy partial fractions decomposition, we get\nThe inverse Laplace transform of\nis\nTherefore the solution to this initial-value problem (IVP) is","attrs_markdown":"[Sign In](https:\/\/auth.fandom.com\/signin?source=mw&redirect=https%3A%2F%2Fmath-physics-problems.fandom.com%2Fwiki%2FLaplace_Transform)\n\n[Register](https:\/\/auth.fandom.com\/register?source=mw&redirect=https%3A%2F%2Fmath-physics-problems.fandom.com%2Fwiki%2FLaplace_Transform)\n\n[![Math & Physics Problems Wikia](https:\/\/static.wikia.nocookie.net\/math-physics-problems\/images\/e\/e6\/Site-logo.png\/revision\/latest?cb=20210601143241)](https:\/\/math-physics-problems.fandom.com\/)   [Math & Physics Problems Wikia](https:\/\/math-physics-problems.fandom.com\/)\n\n- Explore\n  - [Main Page](https:\/\/math-physics-problems.fandom.com\/wiki\/Math_%26_Physics_Problems_Wikia)\n  - 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[Main Page](https:\/\/math-physics-problems.fandom.com\/wiki\/Math_%26_Physics_Problems_Wikia)\n  - [Discuss](https:\/\/math-physics-problems.fandom.com\/f)\n  - [All Pages](https:\/\/math-physics-problems.fandom.com\/wiki\/Special:AllPages)\n  - [Community](https:\/\/math-physics-problems.fandom.com\/wiki\/Special:Community)\n  - [Interactive Maps](https:\/\/math-physics-problems.fandom.com\/wiki\/Special:AllMaps)\n  - [Recent Blog Posts](https:\/\/math-physics-problems.fandom.com\/Blog:Recent_posts)\n- Popular pages\n  - Most visited articles\n    - [Stopped Pipe and Open Pipe](https:\/\/math-physics-problems.fandom.com\/wiki\/Stopped_Pipe_and_Open_Pipe)\n    - [Brahmagupta (ब्रह्मगुप्त)](https:\/\/math-physics-problems.fandom.com\/wiki\/Brahmagupta_\\(%E0%A4%AC%E0%A5%8D%E0%A4%B0%E0%A4%B9%E0%A5%8D%E0%A4%AE%E0%A4%97%E0%A5%81%E0%A4%AA%E0%A5%8D%E0%A4%A4\\))\n    - [Ideological Subversion Interview](https:\/\/math-physics-problems.fandom.com\/wiki\/Ideological_Subversion_Interview)\n    - 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y'- 2y = 4{e}^{-t} } ![{\\\\displaystyle y''-y'-2y=4{e}^{-t}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/ee8f5ab95f3886e4991f20d039521cfcaace1f6a)subject to the initial-values y ( 0 ) \\= 0 {\\\\displaystyle y(0) = 0 } ![{\\\\displaystyle y(0)=0}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/41a64ae0d526d5a958dac00e51adf5d8efeb7cd5)andy ′ ( 0 ) \\= 0\\. {\\\\displaystyle y'(0)=0.} ![{\\\\displaystyle y'(0)=0.}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/245cb45d8b704c1e2d66f526b3c2a1c24426d662)\n\n[![Figure 1](https:\/\/static.wikia.nocookie.net\/math-physics-problems\/images\/b\/b8\/Laplace-Transform-of-Elementary-Functions-1.jpg\/revision\/latest\/scale-to-width-down\/330?cb=20190424155257)](https:\/\/static.wikia.nocookie.net\/math-physics-problems\/images\/b\/b8\/Laplace-Transform-of-Elementary-Functions-1.jpg\/revision\/latest?cb=20190424155257)\n\nFigure 1. Illustrative description of the Laplace transform\n## **Solution**\\[ \\]\n[![Figure 2](https:\/\/static.wikia.nocookie.net\/math-physics-problems\/images\/a\/ae\/Laplacetable-140326134619-phpapp02-thumbnail-4.jpg\/revision\/latest\/scale-to-width-down\/330?cb=20190424155627)](https:\/\/static.wikia.nocookie.net\/math-physics-problems\/images\/a\/ae\/Laplacetable-140326134619-phpapp02-thumbnail-4.jpg\/revision\/latest?cb=20190424155627)\n\nFigure 2. Table of Laplace transforms\n\nCheck the table of Laplace transforms (Figure 1) for the relevant Laplace transformations. The following Laplace transforms will be useful for this differential equation.\n\nL\n\n\\[\n\ny\n\n\\]\n\n\\=\n\nY\n\n(\n\ns\n\n)\n\nL\n\n\\[\n\ny\n\n′\n\n\\]\n\n\\=\n\ns\n\nY\n\n(\n\ns\n\n)\n\n−\n\ny\n\n(\n\n0\n\n)\n\nL\n\n\\[\n\ny\n\n″\n\n\\]\n\n\\=\n\ns\n\n2\n\nY\n\n(\n\ns\n\n)\n\n−\n\ns\n\n⋅\n\ny\n\n(\n\n0\n\n)\n\n−\n\ny\n\n′\n\n(\n\n0\n\n)\n\nL\n\n\\[\n\ne\n\na\n\nt\n\n\\]\n\n\\=\n\n1\n\ns\n\n−\n\na\n\n{\\\\displaystyle {\\\\begin{aligned}\\&L\\[y\\]=Y(s)\\\\\\\\\\[5pt\\]\\&L\\[y'\\]=sY(s)-y(0)\\\\\\\\\\[5pt\\]\\&L\\[y''\\]=s^{2}Y(s)-s\\\\cdot y(0)-y'(0)\\\\\\\\\\[5pt\\]\\&L\\[{e}^{at}\\]={\\\\frac {1}{s-a}}\\\\end{aligned}}}\n\n![{\\\\displaystyle {\\\\begin{aligned}\\&L\\[y\\]=Y(s)\\\\\\\\\\[5pt\\]\\&L\\[y'\\]=sY(s)-y(0)\\\\\\\\\\[5pt\\]\\&L\\[y''\\]=s^{2}Y(s)-s\\\\cdot y(0)-y'(0)\\\\\\\\\\[5pt\\]\\&L\\[{e}^{at}\\]={\\\\frac {1}{s-a}}\\\\end{aligned}}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/3a1be5e35e46ff12bf7a10aba4190928a32671d2)\n\nThis converts the differential equation into the following equation\n\n(\n\ns\n\n2\n\nY\n\n(\n\ns\n\n)\n\n−\n\ns\n\n⋅\n\n0\n\n−\n\n0\n\n)\n\n−\n\n(\n\ns\n\nY\n\n(\n\ns\n\n)\n\n−\n\n0\n\n)\n\n−\n\n2\n\nY\n\n(\n\ns\n\n)\n\n\\=\n\n4\n\ns\n\n\\+\n\n1\n\nY\n\n(\n\ns\n\n)\n\n(\n\ns\n\n2\n\n−\n\ns\n\n−\n\n2\n\n)\n\n\\=\n\n4\n\ns\n\n\\+\n\n1\n\n{\\\\displaystyle {\\\\begin{aligned}\\\\left(s^{2}Y(s)-s\\\\cdot 0-0\\\\right)-\\\\left(sY(s)-0\\\\right)-2Y(s)&={\\\\frac {4}{s+1}}\\\\\\\\\\[5pt\\]Y(s)\\\\left(s^{2}-s-2\\\\right)&={\\\\frac {4}{s+1}}\\\\\\\\\\[5pt\\]\\\\end{aligned}}}\n\n![{\\\\displaystyle {\\\\begin{aligned}\\\\left(s^{2}Y(s)-s\\\\cdot 0-0\\\\right)-\\\\left(sY(s)-0\\\\right)-2Y(s)&={\\\\frac {4}{s+1}}\\\\\\\\\\[5pt\\]Y(s)\\\\left(s^{2}-s-2\\\\right)&={\\\\frac {4}{s+1}}\\\\\\\\\\[5pt\\]\\\\end{aligned}}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/cd46f1c21ca98daf4a45e961d6062d1617bdd662)\n\nSolving for Y ( s ) {\\\\displaystyle Y(s)} ![{\\\\displaystyle Y(s)}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/ee6d07ddc2098d5b3919b899b70ce79e013548c0) givesY ( s ) \\= 4 ( s − 2 ) ( s \\+ 1 ) 2 {\\\\displaystyle Y(s) = \\\\frac{4}{(s-2){(s+1)}^{2}} } ![{\\\\displaystyle Y(s)={\\\\frac {4}{(s-2){(s+1)}^{2}}}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/da7dca04aad5fe28ccb26f1f028ec225775fd05a)\n\nBy partial fractions decomposition, we get\n\nY\n\n(\n\ns\n\n)\n\n\\=\n\n4\n\n\/\n\n9\n\ns\n\n−\n\n2\n\n\\+\n\n−\n\n4\n\n\/\n\n9\n\ns\n\n\\+\n\n1\n\n\\+\n\n−\n\n4\n\n\/\n\n3\n\n(\n\ns\n\n\\+\n\n1\n\n)\n\n2\n\n{\\\\displaystyle Y(s) = \\\\frac{4\/9}{s-2} + \\\\frac{-4\/9}{s+1} + \\\\frac{-4\/3}{{(s+1)}^{2}} }\n\n![{\\\\displaystyle Y(s)={\\\\frac {4\/9}{s-2}}+{\\\\frac {-4\/9}{s+1}}+{\\\\frac {-4\/3}{{(s+1)}^{2}}}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/4369f5a2168af6db8bf4a217934b2d66c377705e)\n\nThe inverse Laplace transform of Y ( s ) {\\\\displaystyle Y(s)} ![{\\\\displaystyle Y(s)}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/ee6d07ddc2098d5b3919b899b70ce79e013548c0) is\n\nL\n\n−\n\n1\n\n\\[\n\nY\n\n(\n\ns\n\n)\n\n\\]\n\n\\=\n\n4\n\n9\n\ne\n\n2\n\nt\n\n−\n\n4\n\n9\n\ne\n\n−\n\nt\n\n−\n\n4\n\n3\n\nt\n\ne\n\n−\n\nt\n\n{\\\\displaystyle {L}^{-1} \\[Y(s)\\] = \\\\frac{4}{9} {e}^{2t} - \\\\frac{4}{9} {e}^{-t} - \\\\frac{4}{3} t {e}^{-t} }\n\n![{\\\\displaystyle {L}^{-1}\\[Y(s)\\]={\\\\frac {4}{9}}{e}^{2t}-{\\\\frac {4}{9}}{e}^{-t}-{\\\\frac {4}{3}}t{e}^{-t}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/0c3cef9f7f0cbcb14b85b9827534d60d2c76b96e)\n\nTherefore the solution to this initial-value problem (IVP) is\n\ny\n\n(\n\nt\n\n)\n\n\\=\n\n4\n\n9\n\ne\n\n2\n\nt\n\n−\n\n4\n\n9\n\ne\n\n−\n\nt\n\n−\n\n4\n\n3\n\nt\n\ne\n\n−\n\nt\n\n{\\\\displaystyle y(t) = \\\\frac{4}{9} {e}^{2t} - \\\\frac{4}{9} {e}^{-t} - \\\\frac{4}{3} t {e}^{-t}}\n\n![{\\\\displaystyle y(t)={\\\\frac {4}{9}}{e}^{2t}-{\\\\frac {4}{9}}{e}^{-t}-{\\\\frac {4}{3}}t{e}^{-t}}](https:\/\/services.fandom.com\/mathoid-facade\/v1\/media\/math\/render\/svg\/d0dc0a8511dcc8f5b39ec6607366fe87cd386a09)\n\nCategories\n\n- [Categories](https:\/\/math-physics-problems.fandom.com\/wiki\/Special:Categories \"Special:Categories\"):\n- [Differential Equations](https:\/\/math-physics-problems.fandom.com\/wiki\/Category:Differential_Equations \"Category:Differential Equations\")\n\nCommunity content is available under [CC-BY-SA](https:\/\/www.fandom.com\/licensing) unless otherwise noted.\n\nAdvertisement\n\n## [![Fandom logo](https:\/\/static.wikia.nocookie.net\/6a181c72-e8bf-419b-b4db-18fd56a0eb60)](https:\/\/www.fandom.com\/ \"Fandom logo\")\n### Explore properties\n- [Fandom](https:\/\/www.fandom.com\/)\n- [Fanatical](https:\/\/www.fanatical.com\/)\n- [GameSpot](https:\/\/www.gamespot.com\/)\n- [Metacritic](https:\/\/www.metacritic.com\/)\n- [TV Guide](https:\/\/www.tvguide.com\/)\n- [Honest Entertainment](https:\/\/www.youtube.com\/user\/screenjunkies)\n### Follow Us\n\n### Overview\n- [What is Fandom?](https:\/\/www.fandom.com\/what-is-fandom)\n- [About](https:\/\/www.fandom.com\/about)\n- [Careers](https:\/\/www.fandom.com\/careers)\n- [Press](https:\/\/www.fandom.com\/press)\n- [Contact](https:\/\/www.fandom.com\/about#contact)\n- [Terms of Use](https:\/\/www.fandom.com\/terms-of-use)\n- [Privacy Policy](https:\/\/www.fandom.com\/privacy-policy)\n- [Digital Services Act](https:\/\/www.fandom.com\/digital-services-act)\n- [Global Sitemap](https:\/\/community.fandom.com\/Sitemap)\n- [Local Sitemap](https:\/\/math-physics-problems.fandom.com\/wiki\/Local_Sitemap)\n\n### Community\n- [Community Central](https:\/\/community.fandom.com\/wiki\/Community_Central)\n- [Support](https:\/\/fandom.zendesk.com\/)\n- [Help](https:\/\/community.fandom.com\/wiki\/Help:Contents)\n\n### Advertise\n- [Media Kit](https:\/\/about.fandom.com\/mediakit)\n- [Contact](https:\/\/about.fandom.com\/mediakit#contact)\n\n### Fandom Apps\nTake your favorite fandoms with you and never miss a beat.\n\n![Fandom App logo](https:\/\/static.wikia.nocookie.net\/6c42ce6a-b205-41f5-82c6-5011721932e7)\n\n- [![Store icon](https:\/\/static.wikia.nocookie.net\/3f46912f-45a8-47f0-ac9d-6dca02c3a5ca)](https:\/\/apps.apple.com\/us\/app\/fandom-videos-news-reviews\/id1230063803)\n- [![Store icon](https:\/\/static.wikia.nocookie.net\/6e8f2e52-ab5a-427f-a836-ab57168161ec)](https:\/\/play.google.com\/store\/apps\/details?id=com.fandom.app&referrer=utm_source%3Dwikia%26utm_medium%3Dglobalfooter)\n\nMath & Physics Problems Wikia is a Fandom Lifestyle Community.\n\nView Mobile Site","attrs_readable_markdown":null,"meta_canonical":null}

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Meta Title	Laplace Transform \| Math & Physics Problems Wikia \| Fandom
Meta Description	Solve the differential equation y ″ − y ′ − 2 y = 4 e − t {\displaystyle y'' - y'- 2y = 4{e}^{-t} } subject to the initial-values y ( 0 ) = 0 {\displaystyle y(0) = 0 } and y ′ ( 0 ) = 0. {\displaystyle y'(0)=0.} Check the table of Laplace transforms (Figure 1) for the relevant Laplace...
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Boilerpipe Text	Problem [ ] Solve the differential equation subject to the initial-values and Figure 1. Illustrative description of the Laplace transform Solution [ ] Figure 2. Table of Laplace transforms Check the table of Laplace transforms (Figure 1) for the relevant Laplace transformations. The following Laplace transforms will be useful for this differential equation. This converts the differential equation into the following equation Solving for gives By partial fractions decomposition, we get The inverse Laplace transform of is Therefore the solution to this initial-value problem (IVP) is
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Illustrative description of the Laplace transform ## Solution**\[ \] [![Figure 2](https://static.wikia.nocookie.net/math-physics-problems/images/a/ae/Laplacetable-140326134619-phpapp02-thumbnail-4.jpg/revision/latest/scale-to-width-down/330?cb=20190424155627)](https://static.wikia.nocookie.net/math-physics-problems/images/a/ae/Laplacetable-140326134619-phpapp02-thumbnail-4.jpg/revision/latest?cb=20190424155627) Figure 2. Table of Laplace transforms Check the table of Laplace transforms (Figure 1) for the relevant Laplace transformations. The following Laplace transforms will be useful for this differential equation. L \[ y \] \= Y ( s ) L \[ y ′ \] \= s Y ( s ) − y ( 0 ) L \[ y ″ \] \= s 2 Y ( s ) − s ⋅ y ( 0 ) − y ′ ( 0 ) L \[ e a t \] \= 1 s − a {\\displaystyle {\\begin{aligned}\&L\[y\]=Y(s)\\\\\[5pt\]\&L\[y'\]=sY(s)-y(0)\\\\\[5pt\]\&L\[y''\]=s^{2}Y(s)-s\\cdot y(0)-y'(0)\\\\\[5pt\]\&L\[{e}^{at}\]={\\frac {1}{s-a}}\\end{aligned}}} ![{\\displaystyle {\\begin{aligned}\&L\[y\]=Y(s)\\\\\[5pt\]\&L\[y'\]=sY(s)-y(0)\\\\\[5pt\]\&L\[y''\]=s^{2}Y(s)-s\\cdot y(0)-y'(0)\\\\\[5pt\]\&L\[{e}^{at}\]={\\frac {1}{s-a}}\\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/3a1be5e35e46ff12bf7a10aba4190928a32671d2) This converts the differential equation into the following equation ( s 2 Y ( s ) − s ⋅ 0 − 0 ) − ( s Y ( s ) − 0 ) − 2 Y ( s ) \= 4 s \+ 1 Y ( s ) ( s 2 − s − 2 ) \= 4 s \+ 1 {\\displaystyle {\\begin{aligned}\\left(s^{2}Y(s)-s\\cdot 0-0\\right)-\\left(sY(s)-0\\right)-2Y(s)&={\\frac {4}{s+1}}\\\\\[5pt\]Y(s)\\left(s^{2}-s-2\\right)&={\\frac {4}{s+1}}\\\\\[5pt\]\\end{aligned}}} ![{\\displaystyle {\\begin{aligned}\\left(s^{2}Y(s)-s\\cdot 0-0\\right)-\\left(sY(s)-0\\right)-2Y(s)&={\\frac {4}{s+1}}\\\\\[5pt\]Y(s)\\left(s^{2}-s-2\\right)&={\\frac {4}{s+1}}\\\\\[5pt\]\\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/cd46f1c21ca98daf4a45e961d6062d1617bdd662) Solving for Y ( s ) {\\displaystyle Y(s)} ![{\\displaystyle Y(s)}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/ee6d07ddc2098d5b3919b899b70ce79e013548c0) givesY ( s ) \= 4 ( s − 2 ) ( s \+ 1 ) 2 {\\displaystyle Y(s) = \\frac{4}{(s-2){(s+1)}^{2}} } ![{\\displaystyle Y(s)={\\frac {4}{(s-2){(s+1)}^{2}}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/da7dca04aad5fe28ccb26f1f028ec225775fd05a) By partial fractions decomposition, we get Y ( s ) \= 4 / 9 s − 2 \+ − 4 / 9 s \+ 1 \+ − 4 / 3 ( s \+ 1 ) 2 {\\displaystyle Y(s) = \\frac{4/9}{s-2} + \\frac{-4/9}{s+1} + \\frac{-4/3}{{(s+1)}^{2}} } ![{\\displaystyle Y(s)={\\frac {4/9}{s-2}}+{\\frac {-4/9}{s+1}}+{\\frac {-4/3}{{(s+1)}^{2}}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/4369f5a2168af6db8bf4a217934b2d66c377705e) The inverse Laplace transform of Y ( s ) {\\displaystyle Y(s)} ![{\\displaystyle Y(s)}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/ee6d07ddc2098d5b3919b899b70ce79e013548c0) is L − 1 \[ Y ( s ) \] \= 4 9 e 2 t − 4 9 e − t − 4 3 t e − t {\\displaystyle {L}^{-1} \[Y(s)\] = \\frac{4}{9} {e}^{2t} - \\frac{4}{9} {e}^{-t} - \\frac{4}{3} t {e}^{-t} } ![{\\displaystyle {L}^{-1}\[Y(s)\]={\\frac {4}{9}}{e}^{2t}-{\\frac {4}{9}}{e}^{-t}-{\\frac {4}{3}}t{e}^{-t}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/0c3cef9f7f0cbcb14b85b9827534d60d2c76b96e) Therefore the solution to this initial-value problem (IVP) is y ( t ) \= 4 9 e 2 t − 4 9 e − t − 4 3 t e − t {\\displaystyle y(t) = \\frac{4}{9} {e}^{2t} - \\frac{4}{9} {e}^{-t} - \\frac{4}{3} t {e}^{-t}} ![{\\displaystyle y(t)={\\frac {4}{9}}{e}^{2t}-{\\frac {4}{9}}{e}^{-t}-{\\frac {4}{3}}t{e}^{-t}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/d0dc0a8511dcc8f5b39ec6607366fe87cd386a09) Categories - [Categories](https://math-physics-problems.fandom.com/wiki/Special:Categories "Special:Categories"): - [Differential Equations](https://math-physics-problems.fandom.com/wiki/Category:Differential_Equations "Category:Differential Equations") Community content is available under [CC-BY-SA](https://www.fandom.com/licensing) unless otherwise noted. 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