Question

Find an explicit solution of the given initial-value problem.

$\left(1+x^4\right)\mathit{d}\mathit{y}\mathbf{+}\mathit{x}\left(1+4y^2\right)\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y=\frac{1-x^2}{2\left(1+x^2\right)}$

Step-by-step solution

Definition of separable equation.

A first-order differential equation of the form

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$

is said to be separable or to have separable variables.

Separate the variables and integration.

$$\begin{gathered} \left(1+x^4\right)dy+x\left(1+4y^2\right)dx=0 \\ \left(1+x^4\right)dy=-x\left(1+4y^2\right)dx \end{gathered}$$

Separating variables

$\frac{1}{\left(1+4y^2\right)}dy=-\frac{x}{\left(1+x^4\right)}dx$

Taking integral on both sides

$\int \frac{1}{\left(1+4y^2\right)}dy=-\int \frac{x}{\left(1+{\left(x^2\right)}^2\right)}dx$

Writing in the form of $\int \frac{1}{\left(a^2+x^2\right)}dx$

$\frac{1}{4}\int \frac{1}{\left(\frac{1}{2^2}+y^2\right)}dy=-\frac{1}{2}\int \frac{2x}{1+{\left(x^2\right)}^2}dx$

Applying formula $\int \frac{1}{\left(a^2+x^2\right)}dx=\frac{1}{a}\arctan \left(\frac{x}{a}\right)+c$

$\arctan \left(2y\right)=-\arctan \left(x^2\right)+c\Rightarrow \left(1\right)$

Substitute the initial condition.

Applying the condition $y\left(1\right)=0$

$0=-\frac{\pi }{4}+c$

After applying the condition we got the value of c

$c=\frac{\pi }{4}$

After putting the value of c into (1)

$2y=\tan \left(\frac{\pi }{4}-\arctan \left(x^2\right)\right)$

Applying the formula $\tan (A-B)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}$

$2y=\frac{\tan \left(\frac{\pi }{4}\right)-\left(\arctan \left(x^2\right)\right)}{1+\tan \left(\arctan \left(x^2\right)\right)\tan \frac{\pi }{4}}$

$y=\frac{1-x^2}{2\left(1+x^2\right)}$