关于用变量替换法求解一阶齐次微分方程的疑问

Hey, great job getting to the integral step—you nailed the substitution ( y = \frac{x}{t} ) for this homogeneous ODE! Let's work through that tricky left-hand integral ( \int \frac{1}{y^3 - 2y} dy ) and finish solving the equation.

Step 1: Factor the Denominator and Partial Fraction Decomposition #

First, factor the denominator to make integration easier:
( y^3 - 2y = y(y^2 - 2) = y(y - \sqrt{2})(y + \sqrt{2}) )

We can decompose the rational function into partial fractions:
[
\frac{1}{y(y^2 - 2)} = \frac{A}{y} + \frac{B}{y - \sqrt{2}} + \frac{C}{y + \sqrt{2}}
]

Multiply both sides by ( y(y^2 - 2) ) to eliminate denominators:
[
1 = A(y^2 - 2) + B(y(y + \sqrt{2})) + C(y(y - \sqrt{2}))
]

Solve for Coefficients ( A, B, C ) #

Expand and group like terms:

• Constant term: ( -2A = 1 ) → ( A = -\frac{1}{2} )
• ( y )-term: ( B\sqrt{2} - C\sqrt{2} = 0 ) → ( B = C )
• ( y^2 )-term: ( A + B + C = 0 )

Substitute ( A = -\frac{1}{2} ) and ( B = C ):
[
-\frac{1}{2} + 2B = 0 \implies B = C = \frac{1}{4}
]

So our decomposed fraction is:
[
\frac{1}{y^3 - 2y} = -\frac{1}{2y} + \frac{1}{4(y - \sqrt{2})} + \frac{1}{4(y + \sqrt{2})}
]

Step 2: Integrate Both Sides #

Now integrate the left-hand side and set equal to ( \int \frac{1}{t} dt ):
[
\int \left( -\frac{1}{2y} + \frac{1}{4(y - \sqrt{2})} + \frac{1}{4(y + \sqrt{2})} \right) dy = \int \frac{1}{t} dt
]

Compute each integral:
[
-\frac{1}{2}\ln|y| + \frac{1}{4}\ln|y - \sqrt{2}| + \frac{1}{4}\ln|y + \sqrt{2}| + C_1 = \ln|t| + C_2
]

Combine constants ( C = C_2 - C_1 ) and simplify the left side using logarithm rules:
[
\frac{1}{4}\ln|(y - \sqrt{2})(y + \sqrt{2})| - \frac{1}{2}\ln|y| = \ln|t| + C
]

Since ( (y - \sqrt{2})(y + \sqrt{2}) = y^2 - 2 ), rewrite as:
[
\frac{1}{4}\ln|y^2 - 2| - \frac{1}{2}\ln|y| = \ln|t| + C
]

Step 3: Combine Logarithms and Solve for ( y ) #

Use exponent rules to move coefficients into the logarithms:
[
\ln\left| \frac{(y^2 - 2)^{1/4}}{y{1/2}} \right| = \ln|t| + C
]

Exponentiate both sides to eliminate logarithms (let ( K = \pm e^C ) to absorb absolute value signs):
[
\frac{(y^2 - 2)^{1/4}}{y{1/2}} = Kt
]

Raise both sides to the 4th power:
[
\frac{y^2 - 2}{y^2} = K^4 t^4
]

Let ( D = K^4 ) (a new constant), then rearrange to solve for ( y^2 ):
[
1 - \frac{2}{y^2} = Dt^4 \implies \frac{2}{y^2} = 1 - Dt^4 \implies y^2 = \frac{2}{1 - Dt^4}
]

Take the square root:
[
y = \pm \sqrt{\frac{2}{1 - Dt^4}}
]

Step 4: Substitute Back ( y = \frac{x}{t} ) #

Replace ( y ) with ( \frac{x}{t} ) to get the general solution:
[
x(t) = \pm t\sqrt{\frac{2}{1 - Dt^4}}
]

Step 5: Apply Initial Condition ( x(1) = 2 ) #

Plug in ( t=1 ), ( x=2 ):
[
2 = \pm 1 \cdot \sqrt{\frac{2}{1 - D(1)^4}}
]

Since the result is positive, we take the positive sign:
[
2 = \sqrt{\frac{2}{1 - D}}
]

Square both sides and solve for ( D ):
[
4 = \frac{2}{1 - D} \implies 4(1 - D) = 2 \implies 1 - D = \frac{1}{2} \implies D = \frac{1}{2}
]

So the particular solution simplifies to:
[
x(t) = t\sqrt{\frac{2}{1 - \frac{1}{2}t^4}} = \frac{2t}{\sqrt{2 - t^4}}
]

内容的提问来源于stack exchange，提问作者George Dimitriou