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

1 Answer

2 votes

Answer:


y = \tan(x + (x^2)/(2))

Explanation:

Poorly formatted question; The complete question requires that we prove that
y=\tan(x+(x\²)/(2))

When


(dy)/(dx) =1+xy\²+x+y\² and
y(0)=0

We have:


(dy)/(dx) =1+xy\²+x+y\²

Rewrite as:


(dy)/(dx) =1+x+xy\²+y\²

Factorize


(dy)/(dx) = (1+x)+y\²(x+1)

Rewrite as:


(dy)/(dx) = (1+x)+y\²(1+x)

Factor out 1 + x


(dy)/(dx) = (1+y\²)(1+x)

Multiply both sides by
(dx)/(1 + y^2)


(dy)/(1+y\²) = (1+x)dx

Integrate both sides


\int (dy)/(1+y\²) = \int (1+x)dx

Rewrite as:


\int (1)/(1+y\²) dy = \int (1+x)dx

Integrate the left-hand side


\int (1)/(1+y\²) dy = \tan^(-1)y

Integrate the right-hand side


\tan^(-1)y = x + (x^2)/(2) + c


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

So:


\tan^(-1)y = x + (x^2)/(2) + c becomes


\tan^(-1)(0) = 0 + (0^2)/(2) + c

This gives:


0 = 0 +0 + c


0 =c


c = 0

The equation
\tan^(-1)y = x + (x^2)/(2) + c becomes


\tan^(-1)y = x + (x^2)/(2) + 0


\tan^(-1)y = x + (x^2)/(2)

Take tan of both sides


y = \tan(x + (x^2)/(2)) --- Proved

User Vignesh Krishnan
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