Question

: Show that the solution of the differential equation: = − − − − − is of the form: + + ( − ) = + , When = and =

Answer 1

Answer:

$y = \tan(x + \frac{x^2}{2})$

Step-by-step explanation:

Poorly formatted question; The complete question requires that we prove that $y=\tan(x+\frac{x\²}{2})$

When

$\frac{dy}{dx} =1+xy\²+x+y\²$ and $y(0)=0$  

We have:

$\frac{dy}{dx} =1+xy\²+x+y\²$

Rewrite as:

$\frac{dy}{dx} =1+x+xy\²+y\²$

Factorize

$\frac{dy}{dx} = (1+x)+y\²(x+1)$

Rewrite as:

$\frac{dy}{dx} = (1+x)+y\²(1+x)$

Factor out 1 + x

$\frac{dy}{dx} = (1+y\²)(1+x)$

Multiply both sides by $\frac{dx}{1 + y^2}$

$\frac{dy}{1+y\²} = (1+x)dx$

Integrate both sides

$\int \frac{dy}{1+y\²} = \int (1+x)dx$

Rewrite as:

$\int \frac{1}{1+y\²} dy = \int (1+x)dx$

Integrate the left-hand side

$\int \frac{1}{1+y\²} dy = \tan^{-1}y$

Integrate the right-hand side

$\tan^{-1}y = x + \frac{x^2}{2} + c$

$y(0)=0$ implies that: $(x,y) = (0,0)$

So:

$\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}(0) = 0 + \frac{0^2}{2} + c$

This gives:

$0 = 0 +0 + c$

$0 =c$

$c = 0$

The equation $\tan^{-1}y = x + \frac{x^2}{2} + c$ becomes

$\tan^{-1}y = x + \frac{x^2}{2} + 0$

$\tan^{-1}y = x + \frac{x^2}{2}$

Take tan of both sides

$y = \tan(x + \frac{x^2}{2})$ --- Proved