Question

1.(30 points) Find the solution of the initial value problem $y' = \frac{(x^3 + 4)(y^2 + y)}{(2y + 1)x^2}$, $y(2) = 1$ Solution: $\frac{dy}{dx} = \frac{x^3 + 4y^2 + y}{x^2(2y + 1)}$ separable eq.

          1.(30 points) Find the solution of the initial value problem
$y' = \frac{(x^3 + 4)(y^2 + y)}{(2y + 1)x^2}$, $y(2) = 1$
Solution: $\frac{dy}{dx} = \frac{x^3 + 4y^2 + y}{x^2(2y + 1)}$ separable eq.

1.(30 points) Find the solution of the initial value problem
y' = ((x^3 + 4)(y^2 + y))/((2y + 1)x^2), y(2) = 1
Solution: (dy)/(dx) = (x^3 + 4y^2 + y)/(x^2(2y + 1)) separable eq.

Added by Brandon B.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Breanna Ollech

07/11/2024

Step 1

$$\frac{2y + 1}{y^2 + y} dy = \frac{x^3}{x^2} dx$$ Show more…

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1.(30 points) Find the solution of the initial value problem $$y' = \frac{(x^3 + 4)(y^2 + y)}{(2y + 1)x^2},$$ y(2) = 1 Solution: $$\frac{dy}{dx} = \frac{x^3 + 4y^2 + y}{x^2(2y + 1)}$$ separable eq.