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Question
Solve `(4x+3y-4)dx+(3x-7y-3)dy=0`
Solution
Given, `(4x+3y-4)dx+(3x-7y-3)dy=0`
∴ `M=(4x+3y-4) and (3x-7y-3)`
Differentiating M by y and N by x, we get,
`(dM)/dy=3` And `(dN)/dx=3`
∴` (dM)/dy=(dn)/dx`
∴ The given equations are exact.
For solution,
`int M dx=int (4x+3y-4)dx`
`int M dx = 2x^2+3xy-4x`
`int" (Term is N free from x)"=int-7y-3 dy`
= `(-7y^2)/2-3y`
∴ The final solution is,
`2x^2+3xy-4x-(7y^2)/2-3y=c`
`4x^2+6xy-8x-7y^2-6y=c`
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