2.1.23 Problem 4 (v)

Internal problem ID [19681]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 3. Solutions of first-order equations. Exercises at page 47
Problem number : 4 (v)
Date solved : Friday, February 13, 2026 at 06:11:24 AM
CAS classification : [_separable]

2.1.23.1 Solved using first_order_ode_separable

0.870 (sec)

Entering first order ode separable solver

\begin{align*} x^{\prime }+2 t x+t x^{4}&=0 \\ \end{align*}

The ode

\begin{equation} x^{\prime } = -t x \left (x^{3}+2\right ) \end{equation}

is separable as it can be written as

\begin{align*} x^{\prime }&= -t x \left (x^{3}+2\right )\\ &= f(t) g(x) \end{align*}

Where

\begin{align*} f(t) &= -t\\ g(x) &= x \left (x^{3}+2\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(x)} \,dx} &= \int { f(t) \,dt} \\ \int { \frac {1}{x \left (x^{3}+2\right )}\,dx} &= \int { -t \,dt} \\ \end{align*}

\[ \frac {\ln \left (x\right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6}=-\frac {t^{2}}{2}+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(x)\) is zero, since we had to divide by this above. Solving \(g(x)=0\) or

\[ x \left (x^{3}+2\right )=0 \]

for \(x\) gives

\begin{align*} x&=0\\ x&=-2^{{1}/{3}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (x\right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6} &= -\frac {t^{2}}{2}+c_1 \\ x &= 0 \\ x &= -2^{{1}/{3}} \\ \end{align*}

Direction field \(x^{\prime }+2 t x+t x^{4} = 0\)	Isoclines for \(x^{\prime }+2 t x+t x^{4} = 0\)

Summary of solutions found

\begin{align*} \frac {\ln \left (x\right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6} &= -\frac {t^{2}}{2}+c_1 \\ x &= 0 \\ x &= -2^{{1}/{3}} \\ \end{align*}

2.1.23.2 Solved using first_order_ode_bernoulli

0.244 (sec)

Entering first order ode bernoulli solver

\begin{align*} x^{\prime }+2 t x+t x^{4}&=0 \\ \end{align*}

In canonical form, the ODE is

\begin{align*} x' &= F(t,x)\\ &= -t \,x^{4}-2 t x \end{align*}

This is a Bernoulli ODE.

\[ x' = \left (-2 t\right ) x + \left (-t\right )x^{4} \tag {1} \]

The standard Bernoulli ODE has the form

\[ x' = f_0(t)x+f_1(t)x^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-2 t\\ f_1 &=-t \end{align*}

The first step is to divide the above equation by \(x^n \) which gives

\[ \frac {x'}{x^n} = f_0(t) x^{1-n} +f_1(t) \tag {3} \]

The next step is use the substitution \(v = x^{1-n}\) in equation (3) which generates a new ODE in \(v \left (t \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(x(t)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(t)&=-2 t\\ f_1(t)&=-t\\ n &=4 \end{align*}

Dividing both sides of ODE (1) by \(x^n=x^{4}\) gives

\begin{align*} x'\frac {1}{x^{4}} &= -\frac {2 t}{x^{3}} -t \tag {4} \end{align*}

Let

\begin{align*} v &= x^{1-n} \\ &= \frac {1}{x^{3}} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(t\) gives

\begin{align*} v' &= -\frac {3}{x^{4}}x' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -\frac {v^{\prime }\left (t \right )}{3}&= -2 v \left (t \right ) t -t\\ v' &= 6 t v +3 t \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (t \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} v^{\prime }\left (t \right ) + q(t)v \left (t \right ) &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=-6 t\\ p(t) &=3 t \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -6 t d t}\\ &= {\mathrm e}^{-3 t^{2}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu v\right ) &= \left (\mu \right ) \left (3 t\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (v \,{\mathrm e}^{-3 t^{2}}\right ) &= \left ({\mathrm e}^{-3 t^{2}}\right ) \left (3 t\right ) \\ \mathrm {d} \left (v \,{\mathrm e}^{-3 t^{2}}\right ) &= \left (3 t \,{\mathrm e}^{-3 t^{2}}\right )\, \mathrm {d} t \\ \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{-3 t^{2}}&= \int {3 t \,{\mathrm e}^{-3 t^{2}} \,dt} \\ &=-\frac {{\mathrm e}^{-3 t^{2}}}{2} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-3 t^{2}}\) gives the final solution

\[ v \left (t \right ) = {\mathrm e}^{3 t^{2}} \left (-\frac {{\mathrm e}^{-3 t^{2}}}{2}+c_1 \right ) \]

The substitution \(v = x^{1-n}\) is now used to convert the above solution back to \(x\) which results in

\[ \frac {1}{x^{3}} = {\mathrm e}^{3 t^{2}} \left (-\frac {{\mathrm e}^{-3 t^{2}}}{2}+c_1 \right ) \]

Simplifying the above gives

\begin{align*} \frac {1}{x^{3}} &= -\frac {1}{2}+c_1 \,{\mathrm e}^{3 t^{2}} \\ \end{align*}

Direction field \(x^{\prime }+2 t x+t x^{4} = 0\)	Isoclines for \(x^{\prime }+2 t x+t x^{4} = 0\)

Summary of solutions found

\begin{align*} \frac {1}{x^{3}} &= -\frac {1}{2}+c_1 \,{\mathrm e}^{3 t^{2}} \\ \end{align*}

2.1.23.3 Solved using first_order_ode_exact

0.309 (sec)

Entering first order ode exact solver

\begin{align*} x^{\prime }+2 t x+t x^{4}&=0 \\ \end{align*}

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(t,x) \mathop {\mathrm {d}t}+ N(t,x) \mathop {\mathrm {d}x}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}x} &= \left (-t \,x^{4}-2 t x\right )\mathop {\mathrm {d}t}\\ \left (t \,x^{4}+2 t x\right ) \mathop {\mathrm {d}t} + \mathop {\mathrm {d}x} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(t,x) &= t \,x^{4}+2 t x\\ N(t,x) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial x} = \frac {\partial N}{\partial t} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial x} &= \frac {\partial }{\partial x} \left (t \,x^{4}+2 t x\right )\\ &= 4 t \,x^{3}+2 t \end{align*}

And

\begin{align*} \frac {\partial N}{\partial t} &= \frac {\partial }{\partial t} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial x} \neq \frac {\partial N}{\partial t}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial x} - \frac {\partial N}{\partial t} \right ) \\ &=1\left ( \left ( 4 t \,x^{3}+2 t\right ) - \left (0 \right ) \right ) \\ &=4 t \,x^{3}+2 t \end{align*}

Since \(A\) depends on \(x\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial t} - \frac {\partial M}{\partial x} \right ) \\ &=\frac {1}{t x \left (x^{3}+2\right )}\left ( \left ( 0\right ) - \left (4 t \,x^{3}+2 t \right ) \right ) \\ &=\frac {-4 x^{3}-2}{x \left (x^{3}+2\right )} \end{align*}

Since \(B\) does not depend on \(t\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}x}} \\ &= e^{\int \frac {-4 x^{3}-2}{x \left (x^{3}+2\right )}\mathop {\mathrm {d}x} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-\ln \left (x \left (x^{3}+2\right )\right ) } \\ &= \frac {1}{x \left (x^{3}+2\right )} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{x \left (x^{3}+2\right )}\left (t \,x^{4}+2 t x\right ) \\ &= t \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{x \left (x^{3}+2\right )}\left (1\right ) \\ &= \frac {1}{x \left (x^{3}+2\right )} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}x}}{\mathop {\mathrm {d}t}} &= 0 \\ \left (t\right ) + \left (\frac {1}{x \left (x^{3}+2\right )}\right ) \frac { \mathop {\mathrm {d}x}}{\mathop {\mathrm {d}t}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (t,x\right )\)

\begin{align*} \frac {\partial \phi }{\partial t } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial x } &= \overline {N}\tag {2} \end{align*}

Integrating (1) w.r.t. \(t\) gives

\begin{align*} \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int \overline {M}\mathop {\mathrm {d}t} \\ \int \frac {\partial \phi }{\partial t} \mathop {\mathrm {d}t} &= \int t\mathop {\mathrm {d}t} \\ \tag{3} \phi &= \frac {t^{2}}{2}+ f(x) \\ \end{align*}

Where \(f(x)\) is used for the constant of integration since \(\phi \) is a function of both \(t\) and \(x\). Taking derivative of equation (3) w.r.t \(x\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial x} = 0+f'(x) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial x} = \frac {1}{x \left (x^{3}+2\right )}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {1}{x \left (x^{3}+2\right )} = 0+f'(x) \end{equation}

Solving equation (5) for \( f'(x)\) gives

\[ f'(x) = \frac {1}{x \left (x^{3}+2\right )} \]

Integrating the above w.r.t \(x\) gives

\begin{align*} \int f'(x) \mathop {\mathrm {d}x} &= \int \left ( \frac {1}{x \left (x^{3}+2\right )}\right ) \mathop {\mathrm {d}x} \\ f(x) &= \frac {\ln \left (x \right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6}+ c_1 \\ \end{align*}

\[ \phi = \frac {t^{2}}{2}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = \frac {t^{2}}{2}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6} \]

Direction field \(x^{\prime }+2 t x+t x^{4} = 0\)	Isoclines for \(x^{\prime }+2 t x+t x^{4} = 0\)

Summary of solutions found

\begin{align*} \frac {\ln \left (x\right )}{2}-\frac {\ln \left (x^{3}+2\right )}{6}+\frac {t^{2}}{2} &= c_1 \\ \end{align*}

2.1.23.4 ✓ Maple. Time used: 0.003 (sec). Leaf size: 123

ode:=diff(x(t),t)+2*x(t)*t+t*x(t)^4 = 0; 
dsolve(ode,x(t), singsol=all);
 

\begin{align*} x &= \frac {2^{{1}/{3}} {\left (\left (2 \,{\mathrm e}^{3 t^{2}} c_1 -1\right )^{2}\right )}^{{1}/{3}}}{2 \,{\mathrm e}^{3 t^{2}} c_1 -1} \\ x &= -\frac {2^{{1}/{3}} {\left (\left (2 \,{\mathrm e}^{3 t^{2}} c_1 -1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{4 \,{\mathrm e}^{3 t^{2}} c_1 -2} \\ x &= \frac {2^{{1}/{3}} {\left (\left (2 \,{\mathrm e}^{3 t^{2}} c_1 -1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}{4 \,{\mathrm e}^{3 t^{2}} c_1 -2} \\ \end{align*}

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d t}x \left (t \right )+2 t x \left (t \right )+t x \left (t \right )^{4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}x \left (t \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}x \left (t \right )=-2 t x \left (t \right )-t x \left (t \right )^{4} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d t}x \left (t \right )}{x \left (t \right ) \left (x \left (t \right )^{3}+2\right )}=-t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {\frac {d}{d t}x \left (t \right )}{x \left (t \right ) \left (x \left (t \right )^{3}+2\right )}d t =\int -t d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (x \left (t \right )^{3}+2\right )}{6}+\frac {\ln \left (x \left (t \right )\right )}{2}=-\frac {t^{2}}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \left (t \right ) \\ {} & {} & \left \{x \left (t \right )=\frac {{\left (-2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )^{2} {\mathrm e}^{-3 t^{2}+6 \mathit {C1}}\right )}^{{1}/{3}}}{{\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1}, x \left (t \right )=-\frac {{\left (-2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )^{2} {\mathrm e}^{-3 t^{2}+6 \mathit {C1}}\right )}^{{1}/{3}}}{2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )}-\frac {\mathrm {I} \sqrt {3}\, {\left (-2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )^{2} {\mathrm e}^{-3 t^{2}+6 \mathit {C1}}\right )}^{{1}/{3}}}{2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )}, x \left (t \right )=-\frac {{\left (-2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )^{2} {\mathrm e}^{-3 t^{2}+6 \mathit {C1}}\right )}^{{1}/{3}}}{2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )}+\frac {\mathrm {I} \sqrt {3}\, {\left (-2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )^{2} {\mathrm e}^{-3 t^{2}+6 \mathit {C1}}\right )}^{{1}/{3}}}{2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )}\right \} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \left \{x \left (t \right )=\frac {{\left (-2 \left ({\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1\right )^{2} {\mathrm e}^{-3 t^{2}+6 \mathit {C1}}\right )}^{{1}/{3}}}{{\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1}, x \left (t \right )=\frac {\left (-1-\mathrm {I} \sqrt {3}\right ) {\left (-\sinh \left (-\frac {3 t^{2}}{2}+3 \mathit {C1} \right )^{2} {\mathrm e}^{-6 t^{2}+12 \mathit {C1}}\right )}^{{1}/{3}}}{{\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1}, x \left (t \right )=\frac {\left (-1+\mathrm {I} \sqrt {3}\right ) {\left (-\sinh \left (-\frac {3 t^{2}}{2}+3 \mathit {C1} \right )^{2} {\mathrm e}^{-6 t^{2}+12 \mathit {C1}}\right )}^{{1}/{3}}}{{\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1}\right \} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{x \left (t \right )=\frac {{\left (-2 \left (\mathit {C1} \,{\mathrm e}^{-3 t^{2}}-1\right )^{2} \mathit {C1} \,{\mathrm e}^{-3 t^{2}}\right )}^{{1}/{3}}}{\mathit {C1} \,{\mathrm e}^{-3 t^{2}}-1}, x \left (t \right )=\frac {\left (-1-\mathrm {I} \sqrt {3}\right ) {\left (-\sinh \left (-\frac {3 t^{2}}{2}+3 \mathit {C1} \right )^{2} {\mathrm e}^{-6 t^{2}+12 \mathit {C1}}\right )}^{{1}/{3}}}{{\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1}, x \left (t \right )=\frac {\left (-1+\mathrm {I} \sqrt {3}\right ) {\left (-\sinh \left (-\frac {3 t^{2}}{2}+3 \mathit {C1} \right )^{2} {\mathrm e}^{-6 t^{2}+12 \mathit {C1}}\right )}^{{1}/{3}}}{{\mathrm e}^{-3 t^{2}+6 \mathit {C1}}-1}\right \} \end {array} \]

2.1.23.5 ✓ Mathematica. Time used: 10.652 (sec). Leaf size: 177

ode=D[x[t],t]+2*t*x[t]+t*x[t]^4==0; 
ic={}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to -\frac {\sqrt [3]{-2} e^{2 c_1}}{\sqrt [3]{e^{3 t^2}-e^{6 c_1}}}\\ x(t)&\to \frac {\sqrt [3]{2} e^{2 c_1}}{\sqrt [3]{e^{3 t^2}-e^{6 c_1}}}\\ x(t)&\to \frac {(-1)^{2/3} \sqrt [3]{2} e^{2 c_1}}{\sqrt [3]{e^{3 t^2}-e^{6 c_1}}}\\ x(t)&\to 0\\ x(t)&\to \sqrt [3]{-2}\\ x(t)&\to -\sqrt [3]{2}\\ x(t)&\to -(-1)^{2/3} \sqrt [3]{2}\\ x(t)&\to \frac {1-i \sqrt {3}}{2^{2/3}} \end{align*}

2.1.23.6 ✓ Sympy. Time used: 2.113 (sec). Leaf size: 88

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(t*x(t)**4 + 2*t*x(t) + Derivative(x(t), t),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

\[ \left [ x{\left (t \right )} = \sqrt [3]{2} \sqrt [3]{- \frac {C_{1}}{C_{1} - e^{3 t^{2}}}}, \ x{\left (t \right )} = \frac {\sqrt [3]{2} \sqrt [3]{- \frac {C_{1}}{C_{1} - e^{3 t^{2}}}} \left (-1 - \sqrt {3} i\right )}{2}, \ x{\left (t \right )} = \frac {\sqrt [3]{2} \sqrt [3]{- \frac {C_{1}}{C_{1} - e^{3 t^{2}}}} \left (-1 + \sqrt {3} i\right )}{2}\right ] \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=x(t)) 
 
('factorable', 'separable', '1st_exact', 'Bernoulli', '1st_power_series', 'lie_group', 'separable_Integral', '1st_exact_Integral', 'Bernoulli_Integral')