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July 2022
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Решите пожалуйста!!!!
y'- y/xlnx = xlnx
y(e)=e²/2
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IrkaShevko
Verified answer
Решаем однородное ДУ:
y' = y/(xlnx)
y'/y = dx/(xlnx)
lny = ln(C*lnx)
y = C*lnx
решение будем искать в виде: y = C(x)*lnx
y' = C'(x)lnx + C(x)/x
C'(x)lnx + C(x)/x - C(x)/x = xlnx
C'(x)lnx = xlnx
C'(x) = x
C(x) = x²/2 + C₁
y = (x²/2 + C₁)lnx
y(e) = e²/2
e²/2 = (e²/2 + C₁)lne = e²/2 + C₁ => C₁ = 0
y = x²lnx/2
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Answers & Comments
Verified answer
Решаем однородное ДУ:y' = y/(xlnx)
y'/y = dx/(xlnx)
lny = ln(C*lnx)
y = C*lnx
решение будем искать в виде: y = C(x)*lnx
y' = C'(x)lnx + C(x)/x
C'(x)lnx + C(x)/x - C(x)/x = xlnx
C'(x)lnx = xlnx
C'(x) = x
C(x) = x²/2 + C₁
y = (x²/2 + C₁)lnx
y(e) = e²/2
e²/2 = (e²/2 + C₁)lne = e²/2 + C₁ => C₁ = 0
y = x²lnx/2