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The standard form is $y'+P(x)y=Q(x)y^n$. To solve, use the substitution $v = y^{1-n}$ which turns it into a linear differential equation.

Use the substitution $v = y^{-1}$. The equation becomes $v' - v = -1$, which has a solution $v = Ce^{-x} - 1$. Hence, $y = \frac{1}{Ce^{-x} - 1}$.

Substitute $v = y^{-2}$. The transformed equation is $v' + 4v = 2e^{2x}$, which can be solved using an integrating factor to find $v$. Then reverse the substitution for $y$.

Apply the substitution $v = y^{-1}$. The new equation $v' - \frac{4}{x}v = - \frac{2}{x^2}$ can be solved. The final solution for $y$ involves combining constants and applying the initial conditions.

Use the substitution $v = y^{-2}$ to linearize the equation to $xv' + 2v = x^3$. Solve for $v$ using integrating factor or other means, then find $y$ from $v$.

The substitution $v = y^{-3}$ linearizes the equation to $v' - 3\tan(x)v = -3\sin(x)$. After finding $v$, the solution for $y$ is found by reversing the substitution.

Substitute $v = y^{-1}$ which converts the equation to $v' - \frac{v}{x} = -x$. Solve the linear ODE for $v$ and then reverse the substitution to find $y$.

Take $v = y^{-4}$ as the substitution, which turns the equation into $v' - 12v = -6x^2$. Solve for $v$ using an integrating factor, then find $y$ by transforming back from $v$.

Substitute $v = y^3$ to get a linear equation $v' - 3v = -3$. Solve for $v$, and then calculate $y$ given that $v = y^3$.