What is the answer to solvefor x,y^2=e^{2x}+2x+c ?

The answer to solvefor x,y^2=e^{2x}+2x+c is x=-(\W(1/(e^{c-y^2)})-y^2+c)/2

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solvefor x,y^2=e^{2x}+2x+c

solve for

x, y^{2} = e^{2 x} + 2 x + c

x = - \frac{W ( \frac{1}{e ^{c - y^{2}}} ) - y ^{2} + c}{2}

Solution steps

y^{2} = e^{2 x} + 2 x + c

Prepare

y^{2} = e^{2 x} + 2 x + c

for Lambert form:

1 = (y^{2} - 2 x - c) e^{- 2 x}

Rewrite the equation with

- 2 x + y^{2} - c = u

and

x = - \frac{u - y ^{2} + c}{2}

1 = (y^{2} - 2 (- \frac{u - y ^{2} + c}{2}) - c) e^{- 2 (- \frac{u - y ^{2} + c}{2})}

Simplify

1 = e^{u - y^{2} + c} u

Rewrite

1 = e^{u - y^{2} + c} u

in Lambert form:

e^{u} u = \frac{1}{e ^{c - y^{2}}}

Solve

e^{u} u = \frac{1}{e ^{c - y^{2}}} : u = W (\frac{1}{e ^{c - y^{2}}})

u = W (\frac{1}{e ^{c - y^{2}}})

Substitute back

u = - 2 x + y^{2} - c,

solve for

x

Solve

- 2 x + y^{2} - c = W (\frac{1}{e ^{c - y^{2}}}) : x = - \frac{W ( \frac{1}{e ^{c - y^{2}}} ) - y ^{2} + c}{2}

x = - \frac{W ( \frac{1}{e ^{c - y^{2}}} ) - y ^{2} + c}{2}

2 5^{x} + 2 5^{x} = 0.4

45 = 220 (0.5)^{x} + 20

65 = 2 5^{x}

lo g_{e} (e^{x}) = 5

(16)^{3 x} = 2

What is the answer to solvefor x,y^2=e^{2x}+2x+c ?
The answer to solvefor x,y^2=e^{2x}+2x+c is x=-(\W(1/(e^{c-y^2)})-y^2+c)/2