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

Question

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

935 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-16T10:18:27

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

#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))#.