Which equation fits? (Differential Equation)

Calculus Level 2

e y = d x d y ( 1 + 4 x 2 ) e 2 x 2 \large e^y = \frac{dx}{dy}\left(1+4x^2\right)e^{2x^{2}}

Given the equation above, what could a possible y y be?

y = 1 x + 2 x y= \frac 1x + 2x y = ln ⁡ x + 2 x 2 y= \ln x + 2x^2 y = 1 x + 2 x 3 y= \frac 1x + 2x^3 y = ln ⁡ ( 2 x ) + x 2 y= \ln (2x) + x^2

Timothy Cao by Timothy Cao , 21, Canada

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1 solution

e y = d x d y ( 1 + 4 x 2 ) e 2 x 2 e y d y = ( 1 + 4 x 2 ) e 2 x 2 d x ∫ e y d y = ∫ ( 1 + 4 x 2 ) e 2 x 2 d x Note that d d x x e 2 x 2 = ( 1 + 4 x 2 ) e 2 x 2 ⟹ e y = x e 2 x 2 + C where C is the constant of integration. e y = e ln ⁡ x + 2 x 2 + C For C = 0 ⟹ y = ln ⁡ x + 2 x 2 \begin{aligned} e^y & = \frac {dx}{dy}\left(1+4x^2\right)e^{2x^2} \\ e^y\ dy & = \left(1+4x^2\right)e^{2x^2}\ dx \\ \int e^y\ dy & = \int \left(1+4x^2\right)e^{2x^2}\ dx & \small \color{#3D99F6} \text{Note that } \frac d{dx} xe^{2x^2} = \left(1+4x^2\right)e^{2x^2} \\ \implies e^y & = xe^{2x^2} + \color{#3D99F6} C & \small \color{#3D99F6} \text{where } C \text{ is the constant of integration.} \\ e^y & = e^{\ln x+2x^2} + C & \small \color{#3D99F6} \text{For }C = 0 \\ \implies y & = \boxed{\ln x + 2x^2} \end{aligned}

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