Question

这个高数题何解

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Answer 1

y'=p
y''=dy'/dx=(dy'/dy)(dy/dx)=pdp/dy
pdp/dy=e^(2y)
dp^2=2e^(2y)dy
p^2=e^2y+C
p=√e^2y+C 或 p=-√(e^2y+C)
dy/√e^2y+C=dx
x=(-1/√C)ln|√[1+Ce^(-2y)]+√Ce^(-y)|+C1 或
x=(1/√C)ln|√[1+Ce^(-2y)]+√Ce^(-y)|+C1
∫dy/√(e^2y+C)=∫dy/e^y√1+Ce^(-2y)=(-1/√C)∫d(√Ce^-y)/√1+Ce^(-2y)
√Ce^(-y)=tanu
=(-1/√C)∫secu^2du/secu
=(-1/√C)∫secudu=(-1/√C)ln|secu+tanu|+C1
=(-1/√C)ln|√[1+Ce^(-2y)] +√Ce^(-y)|+C1

Answer 2

y''=(y')'=>y'=∫y''dy=∫e^(2y)dy=(1/2)∫e^(2y)d2y=(1/2)e^(2y)
=>y=∫(1/2)e^(2y)dy=(1/4)e^(2y)