# Solve Y D X + X ( 1 − 3 X 2 Y 2 ) D Y = 0 - Applied Mathematics 2

Sum

Solve ydx+x(1-3x^2y^2)dy=0

#### Solution

ydx+x(1-3x^2y^2)dy=0  …………….(1)

Compare the given eqn with Mdx + Ndy=0

∴ M = y  therefore N=x(1-3x^2y^2)

(delM)/(dely)=1 (delN)/(delx)=1-9x^2y^2

(delM)/(dely)!=(delN)/(delx)

Hence the given diff. eqn is not exact.
But the given diff. eqn is in the form of yf(xy)dx + xf(xy)dy = 0

Integrating factor = 1/(Mx-Ny)=1/(xy-xy+3x^3y^3)=1/(3x^3y^3)

Multiply the I.F. to eqn (1),

1/(3x^3y^2)dx+[1/(3x^2y^3)-1/y]dy=0

thereforeM_1=1/(3x^3y^2)      N_1=[1/(3x^2y^3)-1/y]

Now this diff. eqn is exact.
The solution of given diff. eqn is given by,

intMdx+int[N-del/(dely)Mdx]dy=c

intM_1dx=int1/(3x^3y^2)dx=(-1)/(6y^2x^2)

del/(dely)intM_1dx=1/(3x^2y^3

int[N_1-del/(dely)intM_1dx]dy=int[1/(3x^2y^3)-1/y-1/(3x^2y^3)]dy

=int(-1)/ydy=-logy

therefore (-1)/(6y^2x^2)-logy=c

Concept: Equations Reducible to Exact Form by Using Integrating Factors
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