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dy/dx=5^(x+y)
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dy/dx=5^(x+y)
▼优质解答
答案和解析
∵令t=x+y,则dy/dx=dt/dx-1
代入原方程,得dt/dx-1=5^t
==>dt/(5^t+1)=dx
==>5^(-t)dt/[1+5^(-t)]=dx
==>(-1/ln5)d[1+5^(-t)]/[1+5^(-t)]=dx
==>d[1+5^(-t)]/[1+5^(-t)]=-ln5*dx
==>ln[1+5^(-t)]=-xln5+ln│C│ (C是积分常数)
==>1+5^(-t)=C/5^x
==>5^t+1=C*5^(t-x)
==>5^(x+y)+1=C*5^y
==>5^(x+y)=C*5^y-1
∴原方程的通解是5^(x+y)=C*5^y-1 (C是积分常数).
代入原方程,得dt/dx-1=5^t
==>dt/(5^t+1)=dx
==>5^(-t)dt/[1+5^(-t)]=dx
==>(-1/ln5)d[1+5^(-t)]/[1+5^(-t)]=dx
==>d[1+5^(-t)]/[1+5^(-t)]=-ln5*dx
==>ln[1+5^(-t)]=-xln5+ln│C│ (C是积分常数)
==>1+5^(-t)=C/5^x
==>5^t+1=C*5^(t-x)
==>5^(x+y)+1=C*5^y
==>5^(x+y)=C*5^y-1
∴原方程的通解是5^(x+y)=C*5^y-1 (C是积分常数).
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