Question

$\frac{dy}{dx} = (x^3 - 2x \tan^{-1}y)(1+y^2)$

Answer 1

$\tan^{-1}y = \frac{1}{2}(x^2-1) + C e^{-x^2}$

Answer 2

The given differential equation is: $\frac{dy}{dx} = (x^3 - 2x \tan^{-1}y)(1+y^2)$

First, rearrange the equation to isolate terms involving $y$ and $dy$: $\frac{1}{1+y^2} \frac{dy}{dx} = x^3 - 2x \tan^{-1}y$

This form suggests a substitution. Let $v = \tan^{-1}y$. Differentiating $v$ with respect to $x$ using the chain rule: $\frac{dv}{dx} = \frac{dv}{dy} \cdot \frac{dy}{dx}$ Since $v = \tan^{-1}y$, $\frac{dv}{dy} = \frac{1}{1+y^2}$. So, $\frac{dv}{dx} = \frac{1}{1+y^2} \frac{dy}{dx}$.

Substitute $v$ and $\frac{dv}{dx}$ into the rearranged differential equation: $\frac{dv}{dx} = x^3 - 2xv$

Rearrange this into the standard form of a first-order linear differential equation, which is $\frac{dv}{dx} + P(x)v = Q(x)$: $\frac{dv}{dx} + (2x)v = x^3$ Here, $P(x) = 2x$ and $Q(x) = x^3$.

To solve this linear differential equation, we find the integrating factor (IF): $\text{IF} = e^{\int P(x) dx} = e^{\int 2x dx} = e^{x^2}$

The general solution for a linear differential equation is given by: $v \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C$ Substituting the values: $v \cdot e^{x^2} = \int x^3 \cdot e^{x^2} dx + C$

Now, we need to evaluate the integral $\int x^3 e^{x^2} dx$. Let $t = x^2$. Then $dt = 2x dx$, which means $x dx = \frac{1}{2} dt$. The integral becomes: $\int x^2 \cdot e^{x^2} \cdot x dx = \int t \cdot e^t \cdot \frac{1}{2} dt = \frac{1}{2} \int t e^t dt$ Use integration by parts for $\int t e^t dt$. Let $u=t$ and $dv=e^t dt$. Then $du=dt$ and $v=e^t$. $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t = e^t(t-1)$ Substitute this back: $\frac{1}{2} \int t e^t dt = \frac{1}{2} e^t(t-1)$ Now, substitute back $t = x^2$: $\frac{1}{2} e^{x^2}(x^2-1)$

Substitute this result back into the general solution equation: $v \cdot e^{x^2} = \frac{1}{2} e^{x^2}(x^2-1) + C$ Divide both sides by $e^{x^2}$ (since $e^{x^2} \neq 0$): $v = \frac{1}{2}(x^2-1) + C e^{-x^2}$

Finally, substitute back $v = \tan^{-1}y$: $\tan^{-1}y = \frac{1}{2}(x^2-1) + C e^{-x^2}$

This is the general solution to the given differential equation.

Explanation of the solution:

The given differential equation is transformed into a first-order linear differential equation by substituting $v = \tan^{-1}y$. This substitution simplifies the left side of the equation from $\frac{1}{1+y^2}\frac{dy}{dx}$ to $\frac{dv}{dx}$. The resulting linear equation $\frac{dv}{dx} + 2xv = x^3$ is then solved using the integrating factor method. The integrating factor is $e^{\int 2x dx} = e^{x^2}$. The solution is $v \cdot e^{x^2} = \int x^3 e^{x^2} dx + C$. The integral $\int x^3 e^{x^2} dx$ is evaluated using substitution ($t=x^2$) followed by integration by parts, yielding $\frac{1}{2}e^{x^2}(x^2-1)$. Substituting this back and solving for $v$, and then replacing $v$ with $\tan^{-1}y$, gives the final solution $\tan^{-1}y = \frac{1}{2}(x^2-1) + C e^{-x^2}$.