anonymous
  • anonymous
hi, need the FINAL answer for the following questions. need to confirm my own answer. \[t \frac{dx}{dt} -x=te ^{\frac{-x}{t}}, x(2)=4\]\[y ^{''}-3y ^{'}+2y=0, y(0)=1, y(3)=0\]\[y''-4y=e ^{x}\cos x, y(0)=1, y'(0)=2\]\[xy'+y=y ^{2},y(1)=-1\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
what were your answers?
anonymous
  • anonymous
sry busy for presentation for the past few days. the answer for each respective questions i'd obtained are \[y=x \ln (\ln \frac{1}{2}+c ^{2})\]for \[t \frac{dx}{dt} -x=te ^{\frac{-x}{t}}, x(2)=4\]
anonymous
  • anonymous
\[y=\frac{e ^{3}}{e ^{3}-1}e ^{x}+\frac{e ^{3}}{e ^{3}-1}e ^{x}\]for \[y ^{''}-3y ^{'}+2y=0, y(0)=1, y(3)=0\]

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anonymous
  • anonymous
\[y=\frac{9}{8}e ^{2x}+\frac{3}{40}e ^{-2x}-\frac{1}{5}e ^{x}\cos x+\frac{1}{10}e ^{x}\sin x\]for\[y''-4y=e ^{x}\cos x, y(0)=1, y'(0)=2\]
anonymous
  • anonymous
but for the last question \[xy'+y=y ^{2},y(1)=-1\]i reached\[2\tanh ^{-1}(1-2y)=\ln x+c\]any1can guide me through? and check the answers for Q1,2,3

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