Please solve the differential equation in its simplest form.

Question

$$sec^2(y) \frac{dy}{dx} + 2x \tan(y) = x^3$$ I am not able to get rid of that $dy/dx$ . Please help. PS: I need the answer as $\tan(y) =\frac{x^2 - 1}{2}$

Answer 1

Put tan y = t

Then we have,

$t' + 2xt = x^3$

It is first order linear equation.

As in the form y' + P(x).y = Q(x)

Solve it to find solution.

u(x) = $e^{\int p(x) dx}$

u(x) = $e^{\int 2x dx}$

u(x) = $e^{x^2}$

Now t = $\frac 1{u(x)} \int u(x). Q(x) dx$

t = $\frac{1}{e^{x^2}} \int e^{x^2}. x^3 dx$

Now solve $\int e^{x^2}. x^3 dx$

Put $x^2 = z$

2x dx = dz

$x dx = \frac12 dz$

Above equation become,

$\frac 12 \int e^z. z dz$

Use integration by parts and replace values.

Answer 2

Hint. Guided by the answer/by inspection, we should make the substitution $z := \tan y$. Then the equation becomes $$ z' + 2xz = x^3$$

Answer 3

HINT:

$$\sec^2\left[\text{y}\left(x\right)\right]\cdot\text{y}'\left(x\right)+2\cdot x\cdot\tan\left[\text{y}\left(x\right)\right]=x^3$$

Let $\text{A}\left(x,\text{y}\right)=2\cdot x\cdot\tan\left(\text{y}\right)-x^3$ and $\text{B}\left(x,\text{y}\right)=\sec^2\left(\text{y}\right)$, this is not an exact equation, because:

$$\frac{\partial\space\text{A}\left(x,\text{y}\right)}{\partial\space\text{y}}=2\cdot x\cdot\sec^2\left(\text{y}\right)\ne0=\frac{\partial\space\text{B}\left(x,\text{y}\right)}{\partial x}$$

Find an integration factor, $\rho\left(x\right)$ such that:

$$\rho\left(x\right)\cdot\text{A}\left(x,\text{y}\right)+\rho\left(x\right)\cdot\text{B}\left(x,\text{y}\right)\cdot\text{y}'\left(x\right)=0$$

is exact.

This means:

$$\frac{\partial}{\partial\space\text{y}}\left[\rho\left(x\right)\cdot\text{A}\left(x,\text{y}\right)\right]=\frac{\partial}{\partial x}\left[\rho\left(x\right)\cdot\text{B}\left(x,\text{y}\right)\right]\implies2\cdot x\cdot\rho\left(x\right)\cdot\sec^2\left(\text{y}\right)=\sec^2\left(\text{y}\right)\cdot\rho\left(x\right)$$

This gives:

$$\int\frac{\rho'\left(x\right)}{\rho\left(x\right)}\space\text{d}x=\int2\cdot x\space\text{d}x=\ln\left|\rho\left(x\right)\right|=x^2$$