Answer:
[tex]\mathbf{((lnx)(2x^2)) -x^2 +39 =(2y^2 +(\dfrac{4}{3})(8+y^2)^{3/2})}[/tex]
Step-by-step explanation:
Given that:
[tex]x In(x) = y( 1+ \sqrt{8+ y^2}) \ } \ \dfrac{dy}{dx}[/tex]
[tex]xln(x) *dx= (y+y \sqrt{(8+y^2}) \ dy[/tex]
[tex]\int xln(x) dx = \int(y+y\sqrt{(8+y^2})) \ dy[/tex]
[tex]\int lnx (x)dx =(\dfrac{y^2}{2} +\dfrac{1}{3}(8+y^2^{)3/2})[/tex]
[tex](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})[/tex]
[tex]((lnx)(\dfrac{x^2}{2}))-\int(\dfrac{x}{2}) =(\dfrac{y^2}{2} +(\dfrac{1}{3})(8+y^2)^{3/2})[/tex]
[tex]((lnx)(\dfrac{x^2}{2}))-(\dfrac{x^2}{4}) +C= (\dfrac{y^2}{2 }+(\dfrac{1}{3})(8+y^2)^{3/2})[/tex]
For y(1) = 1
[tex]((ln(1))(\dfrac{1^2}{2}))-(\dfrac{1^2}{4}) +C= (\dfrac{1^2}{2 }+(\dfrac{1}{3})(8+1^2)^{3/2})[/tex]
[tex](0)-(\dfrac{1}{4}) +C =(\dfrac{1}{2} +(\dfrac{1}{3})(3)^3)[/tex]
[tex]C=(\dfrac{19}{2}) +(\dfrac{1}{4})[/tex]
[tex]C=\dfrac{39}{4}[/tex]
[tex]((lnx)(\dfrac{x^2}{2}))-(\dfrac{x^2}{4}) +(\dfrac{39}{4}) = (\dfrac{y^2}{2} +(\dfrac{1}{3})(8+y^2)^{3/2})[/tex]
[tex]\mathbf{((lnx)(2x^2)) -x^2 +39 =(2y^2 +(\dfrac{4}{3})(8+y^2)^{3/2})}[/tex]
