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y'''=(y"+1)^(1/2).求通解.
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y'''=(y"+1)^(1/2).求通解.
▼优质解答
答案和解析
∵y'''=√(y''+1) ==>dy''/dx=√(y''+1)
==>dy''/√(y''+1)=dx
==>√(y''+1)=x/2+C1 (C1是任意常数)
==>y''=(x/2+C1)²-1
==>y'=2(x/2+C1)³/3-x+C2 (C2是任意常数)
==>y=(x/2+C1)^4/3-x²/2+C2x+C3 (C3是任意常数)
∴原方程的通解是y=(x/2+C1)^4/3-x²/2+C2x+C3.
==>dy''/√(y''+1)=dx
==>√(y''+1)=x/2+C1 (C1是任意常数)
==>y''=(x/2+C1)²-1
==>y'=2(x/2+C1)³/3-x+C2 (C2是任意常数)
==>y=(x/2+C1)^4/3-x²/2+C2x+C3 (C3是任意常数)
∴原方程的通解是y=(x/2+C1)^4/3-x²/2+C2x+C3.
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