Question

Separable differential equation y’ln^2y+ysqrtx=0 y(0)=e

Answer 1

By applying the theory of separable ordinary differential equations we conclude that the solution of the differential equation $\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$ with y(0) = e is $y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$.

How to solve separable differential equation

In this question we must separate each variable on each side of the equivalence, integrate each side of the expression and find an explicit expression (y = f(x)) if possible.

$\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$

$(\ln y)^{2}\,dy = -y \cdot \sqrt{x}\, dx$

$-\frac{(\ln y)^{2}}{y}\, dy = \sqrt{x} \,dx$

$-\int {\frac{(\ln y)^{2}}{y} } \, dy = \int {\sqrt{x}} \, dx$

If u = ㏑ y and du = dy/y, then:

$-\int {u^{2}\,du } = \int {x^{\frac{1}{2} }} \, dx$

$-\frac{1}{3}\cdot u^{3} = \frac{2\cdot x^{\frac{3}{2} }}{3} + C$

$u^{3} = -2\cdot x^{\frac{3}{2} } + C$

$(\ln y)^{3} = - 2\cdot x^{\frac{3}{2} } + C$

$C = (\ln e)^{3}$

$C = 1$

And finally we get the explicit expression:

$\ln y = \sqrt [3]{-2\cdot x^{\frac{3}{2} }+ 1}$

$y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$

By applying the theory of separable ordinary differential equations we conclude that the solution of the differential equation $\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$ with y(0) = e is $y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$.

To learn more on ordinary differential equations: brainly.com/question/14620493

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