Find the solution of the differential equation that satisfies the given initial condition. x ln(x) = y 1 + 8 + y2 y’, y(1) = 1.

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Mathematics Khánh Gia 5 years 2021-08-23T10:33:54+00:00 2021-08-23T10:33:54+00:00 1 Answers 18 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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Tryphena · Answer 1 · 2021-08-23T10:35:40+00:00

Tryphena

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2021-08-23T10:35:40+00:00 August 23, 2021 at 10:35 am

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Answer:

$\mathbf{((lnx)(2x^2)) -x^2 +39 =(2y^2 +(\dfrac{4}{3})(8+y^2)^{3/2})}$

Step-by-step explanation:

Given that:

$x In(x) = y( 1+ \sqrt{8+ y^2}) \ } \ \dfrac{dy}{dx}$

$xln(x) *dx= (y+y \sqrt{(8+y^2}) \ dy$

$\int xln(x) dx = \int(y+y\sqrt{(8+y^2})) \ dy$

$\int lnx (x)dx =(\dfrac{y^2}{2} +\dfrac{1}{3}(8+y^2^{)3/2})$

$(lnx \dfrac{x^2}{2})-\int((\dfrac{x^2}{2})\times \dfrac{1}{x} =(\dfrac{y^2}{2} +\dfrac{1}{3}(8+y^2)^{3/2})$

$((lnx)(\dfrac{x^2}{2}))-\int(\dfrac{x}{2}) =(\dfrac{y^2}{2} +(\dfrac{1}{3})(8+y^2)^{3/2})$

$((lnx)(\dfrac{x^2}{2}))-(\dfrac{x^2}{4}) +C= (\dfrac{y^2}{2 }+(\dfrac{1}{3})(8+y^2)^{3/2})$

For y(1) = 1

$((ln(1))(\dfrac{1^2}{2}))-(\dfrac{1^2}{4}) +C= (\dfrac{1^2}{2 }+(\dfrac{1}{3})(8+1^2)^{3/2})$

$(0)-(\dfrac{1}{4}) +C =(\dfrac{1}{2} +(\dfrac{1}{3})(3)^3)$

$C=(\dfrac{19}{2}) +(\dfrac{1}{4})$

$C=\dfrac{39}{4}$

$((lnx)(\dfrac{x^2}{2}))-(\dfrac{x^2}{4}) +(\dfrac{39}{4}) = (\dfrac{y^2}{2} +(\dfrac{1}{3})(8+y^2)^{3/2})$

$\mathbf{((lnx)(2x^2)) -x^2 +39 =(2y^2 +(\dfrac{4}{3})(8+y^2)^{3/2})}$