Answer:
[tex]y = \tan(x + \frac{x^2}{2})[/tex]
Step-by-step explanation:
Poorly formatted question; The complete question requires that we prove that [tex]y=\tan(x+\frac{x\²}{2})[/tex]
When
[tex]\frac{dy}{dx} =1+xy\²+x+y\²[/tex] and [tex]y(0)=0[/tex]
We have:
[tex]\frac{dy}{dx} =1+xy\²+x+y\²[/tex]
Rewrite as:
[tex]\frac{dy}{dx} =1+x+xy\²+y\²[/tex]
Factorize
[tex]\frac{dy}{dx} = (1+x)+y\²(x+1)[/tex]
Rewrite as:
[tex]\frac{dy}{dx} = (1+x)+y\²(1+x)[/tex]
Factor out 1 + x
[tex]\frac{dy}{dx} = (1+y\²)(1+x)[/tex]
Multiply both sides by [tex]\frac{dx}{1 + y^2}[/tex]
[tex]\frac{dy}{1+y\²} = (1+x)dx[/tex]
Integrate both sides
[tex]\int \frac{dy}{1+y\²} = \int (1+x)dx[/tex]
Rewrite as:
[tex]\int \frac{1}{1+y\²} dy = \int (1+x)dx[/tex]
Integrate the left-hand side
[tex]\int \frac{1}{1+y\²} dy = \tan^{-1}y[/tex]
Integrate the right-hand side
[tex]\tan^{-1}y = x + \frac{x^2}{2} + c[/tex]
[tex]y(0)=0[/tex] implies that: [tex](x,y) = (0,0)[/tex]
So:
[tex]\tan^{-1}y = x + \frac{x^2}{2} + c[/tex] becomes
[tex]\tan^{-1}(0) = 0 + \frac{0^2}{2} + c[/tex]
This gives:
[tex]0 = 0 +0 + c[/tex]
[tex]0 =c[/tex]
[tex]c = 0[/tex]
The equation [tex]\tan^{-1}y = x + \frac{x^2}{2} + c[/tex] becomes
[tex]\tan^{-1}y = x + \frac{x^2}{2} + 0[/tex]
[tex]\tan^{-1}y = x + \frac{x^2}{2}[/tex]
Take tan of both sides
[tex]y = \tan(x + \frac{x^2}{2})[/tex] --- Proved
