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標題:

LInear 1st order differential

發問:

y' - y tan(x) = sin^2 (x) ; y(0)=1 please use dy/dx +p(x)y=q(x) y=1/R(x) integrate R(x)q(x) dx R(x)= exp (integrate p(x) dx) the answer is y=sec (x) (1/3 sin^3(x)+1) please show how can u get the answer steps by steps, thanks!

最佳解答:

R(x)= exp (integrate p(x) dx) integrate p(x)dx = integrate -tanx dx = integrate -sinx/cosx dx = integrate 1/cosx d(cosx) = ln(cosx) R(x) = exp [ln(cos(x))] = cosx y=1/R(x) integrate R(x)q(x) dx integrate R(x)q(x) dx = integrate cos(x) sin^2(x) dx = integrate sin^2(x) d(sin(x)) = 1/3 sin^3(x) + c so y = sec(x)(1/3 sin^3(x) + c) y(0) = c = 1 finally y = sec(x) (1/3 sin^3(x) +1)

其他解答:

• 急急急,a.math一條軌跡數
• O Notation
• 數學,a+b=c , a+c=4b , 2b+c=-c

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