Linear/Seperable Differential Equations Question (Engineering First Year Calculus)?

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If someone could solve this that would be great, thank you so much :)

1 Answer
Jul 16, 2018

xy=e^(1/2-1/(2x^2))xy=e1212x2.

Explanation:

x^3y'=y-x^2y=y(1-x^2).

:. (y')/y=(1-x^2)/x^3,

i.e., 1/y*dy/dx=1/x^3-x^2/x^3,

or, dy/y=(1/x^3-1/x)dx............."[Separable Variable]".

Integrating, intdy/y=int(1/x^3-1/x)dx-lnc,

:. lny=x^-2/-2-lnx-lnc,

i.e., lny+lnx+lnc=-1/(2x^2),

or, ln(yxc)=-1/(2x^2).

:. yxc=e^(-1/(2x^2))......(ast), is a desired Gen. Soln.

To get the Particular Soln., let us use the Initial Cond. (IC)

y(-1)=-1," i.e., when "x=-1, y=-1.

Using IC in (ast), (-1)(-1)c=e^(-1/2) rArr c=e^(-1/2).

Subst.ing in the GS, we get, yxe^(-1/2)=e^(-1/(2x^2))

or, xy=e^(1/2-1/(2x^2)).