A^{A^{A}} A^{A^{A}} B^{B^{B}} B^{B^{B}} C^{C^{C}} C^{C^{C}} D^{D^{D}} D^{D^{D}} E^{E^{E}} E^{E^{E}} F^{F^{F}} F^{F^{F}} G^{G^{G}} G^{G^{G}} H^{H^{H}} H^{H^{H}} I^{I^{I}} I^{I^{I}} J^{J^{J}} J^{J^{J}} K^{K^{K}} K^{K^{K}} L^{L^{L}} L^{L^{L}} M^{M^{M}} M^{M^{M}} N^{N^{N}} N^{N^{N}} O^{O^{O}} O^{O^{O}} P^{P^{P}} P^{P^{P}} Q^{Q^{Q}} Q^{Q^{Q}} R^{R^{R}} R^{R^{R}} S^{S^{S}} S^{S^{S}} T ^{T ^{T}} T ^{T ^{T}} U^{U^{U}} U^{U^{U}} V^{V^{V}} V^{V^{V}} W^{W^{W}} W^{W^{W}} X^{X^{X}} X^{X^{X}} Y^{Y^{Y}} Y^{Y^{Y}} Z^{Z^{Z}} Z^{Z^{Z}}

∐ ⋁ ⋀ ⨄ ⋂ ⋃ ∫ ∏ ∑ ⨂ ⨁ ⨀ ∮ ⨆ ∫ △∀∃¬⊤⊥∇§¶♣♢♡♠^{∐ ⋁ ⋀ ⨄ ⋂ ⋃ ∫ ∏ ∑ ⨂ ⨁ ⨀ ∮ ⨆ ∫△∀∃¬⊤⊥∇§¶♣♢♡♠^{∐ ⋁ ⋀ ⨄ ⋂ ⋃ ∫ ∏ ∑ ⨂ ⨁ ⨀ ∮ ⨆ ∫ △∀∃¬⊤⊥∇§¶♣♢♡♠}}
∐ ⋁ ⋀ ⨄ ⋂ ⋃ ∫ ∏ ∑ ⨂ ⨁ ⨀ ∮ ⨆ ∫ △∀∃¬⊤⊥∇§¶♣♢♡♠^{∐ ⋁ ⋀ ⨄ ⋂ ⋃ ∫ ∏ ∑ ⨂ ⨁ ⨀ ∮ ⨆ ∫△∀∃¬⊤⊥∇§¶♣♢♡♠^{∐ ⋁ ⋀ ⨄ ⋂ ⋃ ∫ ∏ ∑ ⨂ ⨁ ⨀ ∮ ⨆ ∫ △∀∃¬⊤⊥∇§¶♣♢♡♠}}

∗ − +/ ◁ ▷ △ ▽ ∧ ∨ ∩ ∪ ‡ † ⊓ ⊔ ⊎ ⨿ ⋄ • ≀ ÷ ⊙ ⊘ ⊗ ⊖ ⊕ ∓ ± ∘ ◯ \ · ∗ × ⋆·^{∗−+/◁▷△▽∧∨∩∪‡†⊓⊔⊎⨿⋄•≀÷⊙⊘⊗⊖⊕∓±∘◯\·∗×⋆·^{∗−+/◁▷△▽∧∨∩∪‡†⊓⊔⊎⨿⋄•≀÷⊙⊘⊗⊖⊕∓±∘◯\·∗×⋆·}}
∗ − +/ ◁ ▷ △ ▽ ∧ ∨ ∩ ∪ ‡ † ⊓ ⊔ ⊎ ⨿ ⋄ • ≀ ÷ ⊙ ⊘ ⊗ ⊖ ⊕ ∓ ± ∘ ◯ \ · ∗ × ⋆·^{∗−+/◁▷△▽∧∨∩∪‡†⊓⊔⊎⨿⋄•≀÷⊙⊘⊗⊖⊕∓±∘◯\·∗×⋆·^{∗−+/◁▷△▽∧∨∩∪‡†⊓⊔⊎⨿⋄•≀÷⊙⊘⊗⊖⊕∓±∘◯\·∗×⋆·}}

↼↽⇀⇁⌣⌢≍≡⊆⊇≤≥⪯⪰∼≈⊂⊃≪≫≺≻←→↔≃∈∋⊥̸:=><∝⊑⊒∥|⊣⊢↗↘↖↙⇔⇐⇒^{↼↽⇀⇁⌣⌢≍≡⊆⊇≤≥⪯⪰∼≈⊂⊃≪≫≺≻←→↔≃∈∋⊥̸:=><∝⊑⊒∥|⊣⊢↗↘↖↙⇔⇐⇒^{↼↽⇀⇁⌣⌢≍≡⊆⊇≤≥⪯⪰∼≈⊂⊃≪≫≺≻←→↔≃∈∋⊥̸:=><∝⊑⊒∥|⊣⊢↗↘↖↙⇔⇐⇒}}
↼↽⇀⇁⌣⌢≍≡⊆⊇≤≥⪯⪰∼≈⊂⊃≪≫≺≻←→↔≃∈∋⊥̸:=><∝⊑⊒∥|⊣⊢↗↘↖↙⇔⇐⇒^{↼↽⇀⇁⌣⌢≍≡⊆⊇≤≥⪯⪰∼≈⊂⊃≪≫≺≻←→↔≃∈∋⊥̸:=><∝⊑⊒∥|⊣⊢↗↘↖↙⇔⇐⇒^{↼↽⇀⇁⌣⌢≍≡⊆⊇≤≥⪯⪰∼≈⊂⊃≪≫≺≻←→↔≃∈∋⊥̸:=><∝⊑⊒∥|⊣⊢↗↘↖↙⇔⇐⇒}}

()^{(}^{)}^{(}^{)} ()^{(}^{)}^{(}^{)} []^{[}^{]}^{[}^{]} []^{[}^{]}^{[}^{]} ⌊⌋^{⌊}^{⌋}^{⌊}^{⌋} ⌊⌋^{⌊}^{⌋}^{⌊}^{⌋} ⌈⌉^{⌈}^{⌉}^{⌈}^{⌉} ⌈⌉^{⌈}^{⌉}^{⌈}^{⌉} {}^{{}^{}}^{{}^{}} {}^{{}^{}}^{{}^{}} ⟨⟩^{⟨}^{⟩}^{⟨}^{⟩} ⟨⟩^{⟨}^{⟩}^{⟨}^{⟩} ()()( ) ()()( )

[][][ ] [][][ ] ⌊⌋⌊⌋⌊ ⌋ ⌊⌋⌊⌋⌊ ⌋ ⌈⌉⌈⌉⌈ ⌉ ⌈⌉⌈⌉⌈ ⌉ {}{}{ } {}{}{ } ⟨⟩⟨⟩⟨ ⟩ ⟨⟩⟨⟩⟨ ⟩

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A^{A^{A}} A^{A^{A}} B^{B^{B}} B^{B^{B}} C^{C^{C}} C^{C^{C}} D^{D^{D}} D^{D^{D}} E^{E^{E}} E^{E^{E}} F^{F^{F}} F^{F^{F}} G^{G^{G}} G^{G^{G}} H^{H^{H}} H^{H^{H}} I^{I^{I}} I^{I^{I}} J^{J^{J}} J^{J^{J}} K^{K^{K}} K^{K^{K}} L^{L^{L}} L^{L^{L}} M^{M^{M}} M^{M^{M}} N^{N^{N}} N^{N^{N}} O^{O^{O}} O^{O^{O}} P^{P^{P}} P^{P^{P}} Q^{Q^{Q}} Q^{Q^{Q}} R^{R^{R}} R^{R^{R}} S^{S^{S}} S^{S^{S}} T^{T ^{T}} T^{T ^{T}} U^{U^{U}} U^{U^{U}} V^{V^{V}} V^{V^{V}} W^{W^{W}} W^{W^{W}} X^{X^{X}} X^{X^{X}} Y^{Y^{Y}} Y^{Y^{Y}} Z^{Z^{Z}} Z^{Z^{Z}} A^{A^{A}} A^{A^{A}} B^{B^{B}} B^{B^{B}} C^{C^{C}} C^{C^{C}} D^{D^{D}} D^{D^{D}} E^{E^{E}} E^{E^{E}} F^{F^{F}} F^{F^{F}} G^{G^{G}} G^{G^{G}} H^{H^{H}} H^{H^{H}} I^{I^{I}} I^{I^{I}} J^{J^{J}} J^{J^{J}} K^{K^{K}} K^{K^{K}} L^{L^{L}} L^{L^{L}} M^{M^{M}} M^{M^{M}} N^{N^{N}} N^{N^{N}} O^{O^{O}} O^{O^{O}} P^{P^{P}} P^{P^{P}} Q^{Q^{Q}} Q^{Q^{Q}} R^{R^{R}} R^{R^{R}} S^{S^{S}} S^{S^{S}} T ^{T ^{T}} T ^{T ^{T}} U^{U^{U}} U^{U^{U}} V^{V^{V}} V^{V^{V}} W^{W^{W}} W^{W^{W}} X^{X^{X}} X^{X^{X}} Y^{Y^{Y}} Y^{Y^{Y}} Z^{Z^{Z}} Z^{Z^{Z}} ℜ^{ℜ^{ℜ}} ℜ^{ℜ^{ℜ}} ℑ^{ℑ^{ℑ}} ℑ^{ℑ^{ℑ}} 𝔸^{𝔸^{𝔸}} 𝔸^{𝔸^{𝔸}} 𝔹^{𝔹^{𝔹}} 𝔹^{𝔹^{𝔹}} 𝔻^{𝔻^{𝔻}} 𝔻^{𝔻^{𝔻}} 𝔼^{𝔼^{𝔼}} 𝔼^{𝔼^{𝔼}} 𝔽^{𝔽^{𝔽}} 𝔽^{𝔽^{𝔽}} 𝔾^{𝔾^{𝔾}} 𝔾^{𝔾^{𝔾}} 𝕀^{𝕀^{𝕀}} 𝕀^{𝕀^{𝕀}} 𝕁^{𝕁^{𝕁}} 𝕁^{𝕁^{𝕁}} 𝕂^{𝕂^{𝕂}} 𝕂^{𝕂^{𝕂}} 𝕃^{𝕃^{𝕃}} 𝕃^{𝕃^{𝕃}} 𝕄^{𝕄^{𝕄}} 𝕄^{𝕄^{𝕄}} 𝕆^{𝕆^{𝕆}} 𝕆^{𝕆^{𝕆}} 𝕊^{𝕊^{𝕊}} 𝕊^{𝕊^{𝕊}} 𝕋^{𝕋^{𝕋}} 𝕋^{𝕋^{𝕋}} 𝕌^{𝕌^{𝕌}} 𝕌^{𝕌^{𝕌}} 𝕍^{𝕍^{𝕍}} 𝕍^{𝕍^{𝕍}} 𝕎^{𝕎^{𝕎}} 𝕎^{𝕎^{𝕎}} 𝕏^{𝕏^{𝕏}} 𝕏^{𝕏^{𝕏}} 𝕐^{𝕐^{𝕐}} 𝕐^{𝕐^{𝕐}} 𝕜^{𝕜^{𝕜}} 𝕜^{𝕜^{𝕜}} 𝔄^{𝔄^{𝔄}} 𝔄^{𝔄^{𝔄}} 𝔅^{𝔅^{𝔅}} 𝔅^{𝔅^{𝔅}} ℭ^{ℭ^{ℭ}} ℭ^{ℭ^{ℭ}} 𝔇^{𝔇^{𝔇}} 𝔇^{𝔇^{𝔇}} 𝔈^{𝔈^{𝔈}} 𝔈^{𝔈^{𝔈}} 𝔉^{𝔉^{𝔉}} 𝔉^{𝔉^{𝔉}} 𝔊^{𝔊^{𝔊}} 𝔊^{𝔊^{𝔊}} ℌ^{ℌ^{ℌ}} ℌ^{ℌ^{ℌ}} ℑ^{ℑ^{ℑ}} ℑ^{ℑ^{ℑ}} 𝔍^{𝔍^{𝔍}} 𝔍^{𝔍^{𝔍}} 𝔎^{𝔎^{𝔎}} 𝔎^{𝔎^{𝔎}} 𝔏^{𝔏^{𝔏}} 𝔏^{𝔏^{𝔏}} 𝔐^{𝔐^{𝔐}} 𝔐^{𝔐^{𝔐}} 𝔑^{𝔑^{𝔑}} 𝔑^{𝔑^{𝔑}} 𝔒^{𝔒^{𝔒}} 𝔒^{𝔒^{𝔒}} 𝔓^{𝔓^{𝔓}} 𝔓^{𝔓^{𝔓}} 𝔔^{𝔔^{𝔔}} 𝔔^{𝔔^{𝔔}} ℜ^{ℜ^{ℜ}} ℜ^{ℜ^{ℜ}} 𝔖^{𝔖^{𝔖}} 𝔖^{𝔖^{𝔖}} 𝔗^{𝔗^{𝔗}} 𝔗^{𝔗^{𝔗}} 𝔘^{𝔘^{𝔘}} 𝔘^{𝔘^{𝔘}} 𝔙^{𝔙^{𝔙}} 𝔙^{𝔙^{𝔙}} 𝔚^{𝔚^{𝔚}} 𝔚^{𝔚^{𝔚}} 𝔛^{𝔛^{𝔛}} 𝔛^{𝔛^{𝔛}} 𝔜^{𝔜^{𝔜}} 𝔜^{𝔜^{𝔜}} ℨ^{ℨ^{ℨ}} ℨ^{ℨ^{ℨ}} 𝔞^{𝔞^{𝔞}} 𝔞^{𝔞^{𝔞}} 𝔟^{𝔟^{𝔟}} 𝔟^{𝔟^{𝔟}} 𝔠^{𝔠^{𝔠}} 𝔠^{𝔠^{𝔠}} 𝔡^{𝔡^{𝔡}} 𝔡^{𝔡^{𝔡}} 𝔢^{𝔢^{𝔢}} 𝔢^{𝔢^{𝔢}} 𝔣^{𝔣^{𝔣}} 𝔣^{𝔣^{𝔣}} 𝔤^{𝔤^{𝔤}} 𝔤^{𝔤^{𝔤}} 𝔥^{𝔥^{𝔥}} 𝔥^{𝔥^{𝔥}} 𝔦^{𝔦^{𝔦}} 𝔦^{𝔦^{𝔦}} 𝔧^{𝔧^{𝔧}} 𝔧^{𝔧^{𝔧}} 𝔨^{𝔨^{𝔨}} 𝔨^{𝔨^{𝔨}} 𝔩^{𝔩^{𝔩}} 𝔩^{𝔩^{𝔩}} 𝔪^{𝔪^{𝔪}} 𝔪^{𝔪^{𝔪}} 𝔫^{𝔫^{𝔫}} 𝔫^{𝔫^{𝔫}} 𝔬^{𝔬^{𝔬}} 𝔬^{𝔬^{𝔬}} 𝔭^{𝔭^{𝔭}} 𝔭^{𝔭^{𝔭}} 𝔮^{𝔮^{𝔮}} 𝔮^{𝔮^{𝔮}} 𝔯^{𝔯^{𝔯}} 𝔯^{𝔯^{𝔯}} 𝔰^{𝔰^{𝔰}} 𝔰^{𝔰^{𝔰}} 𝔱^{𝔱^{𝔱}} 𝔱^{𝔱^{𝔱}} 𝔲^{𝔲^{𝔲}} 𝔲^{𝔲^{𝔲}} 𝔳^{𝔳^{𝔳}} 𝔳^{𝔳^{𝔳}} 𝔴^{𝔴^{𝔴}} 𝔴^{𝔴^{𝔴}} 𝔵^{𝔵^{𝔵}} 𝔵^{𝔵^{𝔵}} 𝔶^{𝔶^{𝔶}} 𝔶^{𝔶^{𝔶}} 𝔷^{𝔷^{𝔷}} 𝔷^{𝔷^{𝔷}} 0^{0^{0}} 0^{0^{0}} 1^{1^{1}} 1^{1^{1}} 2^{2^{2}} 2^{2^{2}} 3^{3^{3}} 3^{3^{3}} 4^{4^{4}} 4^{4^{4}} 5^{5^{5}} 5^{5^{5}} 6^{6^{6}} 6^{6^{6}} 7^{7^{7}} 7^{7^{7}} 8^{8^{8}} 8^{8^{8}} 9^{9^{9}} 9^{9^{9}} 𝔄^{𝔄^{𝔄}} 𝔄^{𝔄^{𝔄}} 𝔅^{𝔅^{𝔅}} 𝔅^{𝔅^{𝔅}} ℭ^{ℭ^{ℭ}} ℭ^{ℭ^{ℭ}} 𝔇^{𝔇^{𝔇}} 𝔇^{𝔇^{𝔇}} 𝔈^{𝔈^{𝔈}} 𝔈^{𝔈^{𝔈}} 𝔉^{𝔉^{𝔉}} 𝔉^{𝔉^{𝔉}} 𝔊^{𝔊^{𝔊}} 𝔊^{𝔊^{𝔊}} ℌ^{ℌ^{ℌ}} ℌ^{ℌ^{ℌ}} ℑ^{ℑ^{ℑ}} ℑ^{ℑ^{ℑ}} 𝔍^{𝔍^{𝔍}} 𝔍^{𝔍^{𝔍}} 𝔎^{𝔎^{𝔎}} 𝔎^{𝔎^{𝔎}} 𝔏^{𝔏^{𝔏}} 𝔏^{𝔏^{𝔏}} 𝔐^{𝔐^{𝔐}} 𝔐^{𝔐^{𝔐}} 𝔑^{𝔑^{𝔑}} 𝔑^{𝔑^{𝔑}} 𝔒^{𝔒^{𝔒}} 𝔒^{𝔒^{𝔒}} 𝔓^{𝔓^{𝔓}} 𝔓^{𝔓^{𝔓}} 𝔔^{𝔔^{𝔔}} 𝔔^{𝔔^{𝔔}} ℜ^{ℜ^{ℜ}} ℜ^{ℜ^{ℜ}} 𝔖^{𝔖^{𝔖}} 𝔖^{𝔖^{𝔖}} 𝔗^{𝔗^{𝔗}} 𝔗^{𝔗^{𝔗}} 𝔘^{𝔘^{𝔘}} 𝔘^{𝔘^{𝔘}} 𝔙^{𝔙^{𝔙}} 𝔙^{𝔙^{𝔙}} 𝔚^{𝔚^{𝔚}} 𝔚^{𝔚^{𝔚}} 𝔛^{𝔛^{𝔛}} 𝔛^{𝔛^{𝔛}} 𝔜^{𝔜^{𝔜}} 𝔜^{𝔜^{𝔜}} ℨ^{ℨ^{ℨ}} ℨ^{ℨ^{ℨ}} 𝔞^{𝔞^{𝔞}} 𝔞^{𝔞^{𝔞}} 𝔟^{𝔟^{𝔟}} 𝔟^{𝔟^{𝔟}} 𝔠^{𝔠^{𝔠}} 𝔠^{𝔠^{𝔠}} 𝔡^{𝔡^{𝔡}} 𝔡^{𝔡^{𝔡}} 𝔢^{𝔢^{𝔢}} 𝔢^{𝔢^{𝔢}} 𝔣^{𝔣^{𝔣}} 𝔣^{𝔣^{𝔣}} 𝔤^{𝔤^{𝔤}} 𝔤^{𝔤^{𝔤}} 𝔥^{𝔥^{𝔥}} 𝔥^{𝔥^{𝔥}} 𝔦^{𝔦^{𝔦}} 𝔦^{𝔦^{𝔦}} 𝔧^{𝔧^{𝔧}} 𝔧^{𝔧^{𝔧}} 𝔨^{𝔨^{𝔨}} 𝔨^{𝔨^{𝔨}} 𝔩^{𝔩^{𝔩}} 𝔩^{𝔩^{𝔩}} 𝔪^{𝔪^{𝔪}} 𝔪^{𝔪^{𝔪}} 𝔫^{𝔫^{𝔫}} 𝔫^{𝔫^{𝔫}} 𝔬^{𝔬^{𝔬}} 𝔬^{𝔬^{𝔬}} 𝔭^{𝔭^{𝔭}} 𝔭^{𝔭^{𝔭}} 𝔮^{𝔮^{𝔮}} 𝔮^{𝔮^{𝔮}} 𝔯^{𝔯^{𝔯}} 𝔯^{𝔯^{𝔯}} 𝔰^{𝔰^{𝔰}} 𝔰^{𝔰^{𝔰}} 𝔱^{𝔱^{𝔱}} 𝔱^{𝔱^{𝔱}} 𝔲^{𝔲^{𝔲}} 𝔲^{𝔲^{𝔲}} 𝔳^{𝔳^{𝔳}} 𝔳^{𝔳^{𝔳}} 𝔴^{𝔴^{𝔴}} 𝔴^{𝔴^{𝔴}} 𝔵^{𝔵^{𝔵}} 𝔵^{𝔵^{𝔵}} 𝔶^{𝔶^{𝔶}} 𝔶^{𝔶^{𝔶}} 𝔷^{𝔷^{𝔷}} 𝔷^{𝔷^{𝔷}}

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Solutions to Homework 2

Math.x280, Spring 2008

Problem 1. A cup of coffee in a room at temperature of 70^{∘} F is being cooled according to Newton’s law of cooling:

\frac{dT}{dt} = −k(T − 70), T (0) = 200. (time is measured in minutes)  (1)

The solution to (1) is

T (t) = 70 + (200 − 70) exp(−kt).  (2)

The coefficient k can be computed from (2) using the fact that T (3) = 175 . Thus,

175 = 70 + 130 exp(−3k)x exp(−k) = \frac{175 − 70}{130} ^{1/3} = \frac{21}{26} ^{1/3}  (3)

The temperature of the coffee will reach 112^{∘} F at t = t_{1} that satisfies the equation

112 = T (t_{1}) = 70 + 130 exp(−kt_{1}) = 70 + 130 exp(−k) ^{t_{1}}, with exp(−k) given in (3)

Thus,

\frac{112 − 70}{130} = \frac{21}{26} ^{1/3} ^{t_{1}}, ort_{1} = \frac{3 log \frac{21}{65}}{log \frac{21}{26}} ≈ 15.87 minutes.

Problem 2. Let N(t) denote the amount of C-14 at time t , with time t = 0 corresponding to the moment when the original amount N_{0} of C-14 was present. Time evolution of N(t) is governed by the equation

\frac{dN}{dt} = −λN, N(0) = N_{0}.  (4)

The amount of C-14 decays in time, thus λ > 0 . From (4), N(t) = N_{0} exp(λt) and, since the half-life of C-14 is approximately 5568 years, N_{0}/2 = N_{0} exp(5568λ) . This gives us λ = − ln 2/5568 ≈ −0.00012448 . Therefore, the expression for N(t) for all time is given by

N(t) = N_{0} exp −\frac{ln 2}{5568}t = N_{0}2^{− \frac{t}{5568}} .  (5)

At present time t , 14.5% is left, or equivalently from (5),

0.145N_{0} = N_{0}2^{− \frac{t}{5568}} x ln 0.145 = −\frac{t}{5568} ln 2xt = −\frac{5568 ln 0.145}{ln 2} ≈ 15511.753 years.

Problem 3. For the equation

(1 + x^{2})dy + (xy + x^{3} + x)dx = 0 or\frac{dy}{dx} + \frac{x}{1 + x^{2}} y = −x,  (6)

an integrating factor is exp ∫ \frac{x}{1 + x^{2}} dx = \sqrt{1 + x^{2}}

so that

\frac{d}{dx} y\sqrt{1 + x^{2}} = −x\sqrt{1 + x^{2}}xy = −\frac{1}{3}(1 + x^{2}) + c 1 + x^{2} ^{− \frac{1}{2}} .  (7)

PIC Figurex1: Graphs of y = −\frac{1}{3}(1 + x^{2}) + c 1 + x^{2} ^{− \frac{1}{2}} for various choices of c .

Problem 4.

(a) For the equation

\frac{3y^{2} − t^{2}}{y^{5}} \frac{dy}{dt} + \frac{t}{2y^{4}} = 0, y(1) = 1,  (8)

let M = t/(2y^{4}) and N = (3y^{2} − t^{2})/y^{5} so that M_{y} = −2t/y^{5} = N_{t} . From ϕ_{t}(t, y) = M = t/(2y^{4}) we obtain ϕ(t, y) = t^{2}/(4y^{4}) + h(y) . Thus, using ϕ_{y} = N = (3y^{2} − t^{2})/y^{5} , we have h′(y) = 3/y^{3} , and h(y) = −3/(2y^{2}) . The general solution is t^{2}/(4y^{4}) − 3/(2y^{2}) = c . If y(1) = 1 then c = −5/4 and the solution to initial-value problem (8) is

\frac{t^{2}}{4y^{4}} − \frac{3}{2y^{2}} = −\frac{5}{4}.  (9)

PIC Figurex2: Graph of the integral curve (9)

PIC

Figurex3: Other integral curves of t^{2}/(4y^{4}) − 3/(2y^{2}) = c

The solutions of the initial value problem (8) and some other integral (dashed) curves of the differential equation

\frac{3y^{2} − t^{2}}{y^{5}} \frac{dy}{dt} + \frac{t}{2y^{4}} = 0 are shown in Figure 3.

Remark 1. Please observe that the graph in Figure 2 does not represent a function !!! Which branch of the graph represents the solution satisfying the initial condition y(1) = 1 .

Remark 2. Note that if one multiplies both sides of equation (8) by y^{5} , the resulting equation

(3y^{2} − t^{2})\frac{dy}{dt} + \frac{1}{2}ty = 0  (10)

is not exact !!! In other words 1/y^{5} is an integrating factor for (10). (See also Problem 5, below)

(b) The initial-value problem

\frac{dy}{dt} = \frac{y}{2y^{2} − t}, y(1) = 5, or\frac{dt}{dy} − \frac{t}{y} = 2y, y(1) = 5,  (11)

is linear in t (as a function of y ). Its integrating factor is 1/y , thus

\frac{d}{dy} \frac{t}{y} = 2xt = 2y^{2} + cy.

If y(1) = 5 then c = −49/5 and t = 2y^{2} − (49/5)y .

PIC

Figurex4: Graph of the integral curve t = 2y^{2} − (49/5)y

PIC

Figurex5: Other integral curves of t = 2y^{2} + cy

\frac{dy}{dt} = \frac{y}{2y^{2} − t}, are shown in Figure 5.

Remark 3. Please observe that the graph in Figure 4 does not represent a function !!! Which branch of the graph represents the solution satisfying the initial condition y(1) = 5 .

Problem 5. Multiply the equation

(t^{2} + ty)\frac{dy}{dt} − y^{2} = 0 by μ(t, y) = \frac{1}{t^{2}y}. The result is  (12)

\frac{1}{y} + \frac{1}{t} \frac{dy}{dt} − \frac{y}{t^{2}} = 0.  (13)

With M = −y/t^{2} and N = 1/y + 1/t we have M_{y} = −1/t^{2} = N_{t} ; therefore equation

(13) is exact. From ϕ_{t} = −y/t^{2} we obtain ϕ(t, y) = y/t + h(y) But ϕ_{y} = N ; thus h′(y) = 1/y , and h(y) = log |y| . The solution of (12) is

ϕ(t, y) = \frac{y}{t} + log |y| = c, where c ∈ ℝ is a constant.

Problem 6. Under the change of the independent variable, t → y , the original equation are transformed into

m\frac{dv}{dt} = −mg − kv^{2}, k > 0 for the “up” motion
m\frac{dv}{dt} = mg − kv^{2}, k > 0 for the “down” motion

x

mv \frac{dv}{dy} = −mg − kv^{2}, k > 0 for the “up” motion
mv \frac{dv}{dy} = mg − kv^{2}, k > 0 for the “down” motion

(14)

Separating variables and integrating the differential equation corresponding to the “up” motion we have

∫ \frac{mv}{mg + kv^{2}} = − ∫ dyx\frac{m}{2k} ln(mg + kv^{2}) = −y + c_{1}xmg + kv^{2} = c_{2} exp −\frac{2ky}{m}

xv^{2} = c_{3} exp −\frac{2ky}{m} − \frac{mg}{k}

(15)

Using y(0) = 0 and v(0) = v_{0} we have that v = v_{0} when y = 0 (the initial condition) so that v_{0}^{2} = c_{3} − mg/k and c_{3} = v_{0}^{2} + mg/k . Thus,

v^{2} = \frac{kv_{0}^{2} + mg}{k} exp −\frac{2ky}{m} − \frac{mg}{k}  (16)

Setting v = 0 in the left hand side of (16) and solving for y we see that the maximum height is

h = \frac{m}{2k} ln \frac{kv_{0}^{2} + mg}{mg} .  (17)

Now, separating variables and integrating the differential equation corresponding to the “down” motion (see (14)) we have

∫ \frac{mv}{mg − kv^{2}} = ∫ dyx − \frac{m}{2k} ln |mg − kv^{2}| = y + d_{1}xmg − kv^{2} = d_{2} exp −\frac{2ky}{m}

xv^{2} = \frac{mg}{k} 1 − d_{3} exp −\frac{2ky}{m}

(18)

In this case v = 0 when y = 0 (the initial condition) so d_{3} = 1 and

v^{2} = \frac{mg}{k} 1 − exp −\frac{2ky}{m} .  (19)

Setting y = h from (17) into (19) and solving for v we obtain that the impact velocity is

v_{i} = \frac{v_{0}}{\sqrt{1 + \frac{kv_{0}^{2}}{mg}}} < v_{0}.  (20)

Remark 4.

By taking y → ∞ in the right hand side of (19) one obtains the limiting velocity, v_{final} , for the problem m dv/dt = mg − kv^{2}

v_{final}^{2} = lim_{y→∞} \frac{mg}{k} 1 − exp −\frac{2ky}{m} = \frac{mg}{k} xv_{final} = \sqrt{\frac{mg}{k}} > \sqrt{\frac{mg}{k}} \sqrt{\frac{v_{0}^{2}}{v_{0}^{2} + \frac{mg}{k}}} = v_{i}. (see, (20))

See also (29), where the same result can be obtained by taking t → ∞ in (29) and using lim_{x→∞} tanh(x) = 1 .

(b) It is interesting to compare the results of Problem 6 with the same problem but without air resistance. When k = 0 the change variables from t to y is not necessary, however, for comparison with the case kx0 , I proceed in an exactly the same way as before. The corresponding equations are

m\frac{dv}{dt} = −mg, for the “up” motion
m\frac{dv}{dt} = mg, for the “down” motion

x

mv \frac{dv}{dy} = −mg, for the “up” motion
mv \frac{dv}{dy} = mg, for the “down” motion

x

(21)

\frac{v^{2}}{2} = −gy + c_{1}, v|_{y=0} = v_{0} for the “up” motion
\frac{v^{2}}{2} = gy + c_{2}v|_{y=0} = 0 for the “down” motion

x

\frac{v^{2}}{2} = −gy + \frac{v_{0}^{2}}{2} , for the “up” motion
\frac{v^{2}}{2} = gy for the “down” motion

Thus, the maximum height is v_{0}^{2}/2g . Observe that the maximum height given in (17) is less than the maximum height in the case without air resistance. Indeed,

\frac{m}{2k} ln \frac{kv_{0}^{2} + mg}{mg} = \frac{m}{2k} ln 1 + \frac{kv_{0}^{2}}{mg} < \frac{m}{2k} \frac{kv_{0}^{2}}{mg} since for x > 0 ln(1 + x) < x = \frac{v_{0}^{2}}{2g}

Also the impact velocity in the case without air resistance is equal to the initial velocity v_{0} . indeed, substituting v_{0}^{2}/2g into v^{2}/g = gy (expression for v for the “down” part of the motion, see, (22)) we obtain

v_{i} = v_{0} (No dissipation of the energy)

Additional remarks to Problem 6

Finally, I show how one can integrate the equations

m\frac{dv}{dt}  = −mg − kv^{2}, v(0) = v_{0}, k > 0 for the “up” motion  (23)
m\frac{dv}{dt}  = mg − kv^{2}, v(0) = 0, k > 0 for the “down” motion  (24)

without changing the variable t to y . The integrations are slightly more involved. For the “up” motion we have, after separation of variables

\frac{m}{k} \frac{dv}{\frac{mg}{k} + v^{2}} = −dtx\frac{1}{\sqrt{\frac{mg}{k}}} arctan v/\sqrt{\frac{mg}{k}} = −\frac{k}{m}t + cxv(t) = \sqrt{\frac{mg}{k}} tan c_{1} − \sqrt{\frac{kg}{m}} t  (25)

Applying the initial conditions v(0) = v_{0} we have

v_{0} = \sqrt{\frac{mg}{k}} tan c_{1}xc_{1} = arctan v_{0}/\sqrt{\frac{mg}{k}} xv(t) = \sqrt{\frac{mg}{k}} tan arctan v_{0}/\sqrt{\frac{mg}{k}} − \sqrt{\frac{kg}{m}} t .  (26)

From (26) the time t_{up} needed for the projectile to attain its maximum height (v(t_{up}) = 0 ) is

t_{up} = \sqrt{\frac{m}{kg}} arctan v_{0}/\sqrt{\frac{mg}{k}}  (27)

For the “down” motion

\frac{m}{k} \frac{dv}{\frac{mg}{k} − v^{2}} = dtx\frac{1}{\sqrt{\frac{mg}{k}}} tanh^{−1} v/\sqrt{\frac{mg}{k}} = \frac{k}{m}t + d_{1}xv(t) = \sqrt{\frac{mg}{k}} tanh d_{1} + \sqrt{\frac{kg}{m}} t  (28)

Applying the initial condition v(0) = 0 one obtains d_{1} = 0 and

v(t) = \sqrt{\frac{mg}{k}} tanh \sqrt{\frac{kg}{m}} t  (29)

The time t_{down} needed for the projectile to fall from the height (17) can also be computed; from (20)

\frac{v_{0}}{\sqrt{1 + \frac{kv_{0}^{2}}{mg}}} = \sqrt{\frac{mg}{k}} tanh \sqrt{\frac{kg}{m}} t x\frac{v_{0}\sqrt{k}}{\sqrt{mg + kv_{0}^{2}}} = tanh \sqrt{\frac{kg}{m}} t

xt_{down} = \sqrt{\frac{mk}{g}} tanh^{−1} \frac{v_{0}\sqrt{k}}{\sqrt{mg + kv_{0}^{2}}}

(30)

One can also show that

t_{up} < t_{down}.  (31)

Furthermore, when air resistance is neglected

t_{up} = t_{down} = \frac{v_{0}}{g} . (No dissipation of the energy)

Problem 7.

(a) The equation for velocity at any time and position y is

v^{2} = v_{0}^{2} − 2gR + \frac{2gR^{2}}{R + y} ,  (32)

where R is the radius of the earth, g is the acceleration due to gravity, and v_{0} is the initial velocity of the projectile (in our case, v_{0} < \sqrt{2gR} ). Now the maximum distance, y = Y_{max} , is reached by the projectile when v = 0 . In other words, we have the following equation for Y_{max} :

0 = V_{0}^{2} − 2gR + \frac{2gR^{2}}{R + Y_{max}} .  (33)

Its solution is

Y_{max} = \frac{Rv_{0}^{2}}{2gR − v_{0}^{2}} .

(b) The solution to the equation y′′ = −g with the initial conditions y(0) = 0 and y′(0) = v(0) = v_{0} > 0 is

y(t) = −\frac{1}{2}gt^{2} + v_{0}t (with

v(t) = −gt + v_{0} ). As in part (a), the maximum distance, y_{max} , is reached when v = 0 ; this corresponds to t = v_{0}/g . Thus, the maximum distance y_{max} is

y_{max} = y \frac{v_{0}}{g} = −\frac{1}{2}g \frac{v_{0}}{g} ^{2} + v_{0} \frac{v_{0}}{g} = \frac{v_{0}^{2}}{2g} .

Finally, since v_{0} < \sqrt{2gR} , we have

Y_{max} = \frac{Rv_{0}^{2}}{2gR − v_{0}^{2}} = \frac{v_{0}^{2}}{2g − v_{0}^{2}/R} > \frac{v_{0}^{2}}{2g} = y_{max}.

(c) In this case (32) becomes

v = \frac{dy}{dt} = \sqrt{\frac{2gR^{2}}{R + y}} , y(0) = 0, Note the plus sign in front of \sqrt{\frac{2gR^{2}}{R + y}}  (34)

which is a separable equation. The solution of (34) is

\frac{2}{3} R + y ^{3/2} = Rt\sqrt{2g} + c.

With y(0) = 0 we have c = 2/3 R^{3/2} . Thus

y(t) = \frac{3\sqrt{2g}}{2} Rt + R^{3/2} ^{2/3} − R.  (35)

(d) Solving (35) for t we obtain

t = \frac{2}{3\sqrt{2g}R} [y(t) + R]^{3/2} − R^{3/2} .  (36)

With R = 3963 miles, y(T_{M}) = 238, 855 miles, and g = 32/5280 = 0.0061 mile/s^{2} , we get from (36) T_{M} = 181853 seconds = 50.52 hours.