hi, need the FINAL answer for the following questions. need to confirm my own answer.
$$t \frac{dx}{dt} -x=te ^{\frac{-x}{t}}, x(2)=4$$$$y ^{''}-3y ^{'}+2y=0, y(0)=1, y(3)=0$$$$y''-4y=e ^{x}\cos x, y(0)=1, y'(0)=2$$$$xy'+y=y ^{2},y(1)=-1$$

Question

hi, need the FINAL answer for the following questions. need to confirm my own answer. 
$$t \frac{dx}{dt} -x=te ^{\frac{-x}{t}}, x(2)=4$$$$y ^{''}-3y ^{'}+2y=0, y(0)=1, y(3)=0$$$$y''-4y=e ^{x}\cos x, y(0)=1, y'(0)=2$$$$xy'+y=y ^{2},y(1)=-1$$

turingtest · Answer

what were your answers?

anonymous · Answer

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 · Answer

$$y=\frac{e ^{3}}{e ^{3}-1}e ^{x}+\frac{e ^{3}}{e ^{3}-1}e ^{x}$$for $$y ^{''}-3y ^{'}+2y=0, y(0)=1, y(3)=0$$

anonymous · Answer

$$y=\frac{9}{8}e ^{2x}+\frac{3}{40}e ^{-2x}-\frac{1}{5}e ^{x}\cos x+\frac{1}{10}e ^{x}\sin x$$for$$y''-4y=e ^{x}\cos x, y(0)=1, y'(0)=2$$

anonymous · Answer

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