Question

The solution of the following IVP $y' = \frac{2}{xy}$, $y(1) = -2$ is given by: $\circ \quad y^2 = 2\ln|x| + 4$ $\circ \quad y = 2\sqrt{4\ln|x|} + 2$ $\circ \quad y = 2\sqrt{\ln|x|} + 1$ $\circ \quad y = -2\sqrt{\ln|x|} + 1$

          The solution of the following IVP
$y' = \frac{2}{xy}$, $y(1) = -2$
is given by:
$\circ \quad y^2 = 2\ln|x| + 4$
$\circ \quad y = 2\sqrt{4\ln|x|} + 2$
$\circ \quad y = 2\sqrt{\ln|x|} + 1$
$\circ \quad y = -2\sqrt{\ln|x|} + 1$

The solution of the following IVP
y' = (2)/(xy), y(1) = -2
is given by:
∘ y^2 = 2ln|x| + 4
∘ y = 2√(4ln|x|) + 2
∘ y = 2√(ln|x|) + 1
∘ y = -2√(ln|x|) + 1

Added by Albert R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Jacob Fry

Step 1

The equation is given as 2y1 = -2, which can be simplified to y1 = -1. Now, let's solve the differential equation. We can do this by integrating both sides of the equation with respect to x. ∫dy1 = ∫-1 dx Integrating both sides gives us: y = -x + C where C Show more…

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