Solve : (2xy cos x
2
– 2xy + 1) dx + (sin x
2
– x
2
) dy = 0
Answers
Answer:
The solution is
\sf{(2xy \cos {x}^{2} - 2xy + 1)dx + ( \sin {x}^{2} - {x}^{2} )dy = 0}(2xycosx
2
−2xy+1)dx+(sinx
2
−x
2
)dy=0
EVALUATION
\sf{(2xy \cos {x}^{2} - 2xy + 1)dx + ( \sin {x}^{2} - {x}^{2} )dy = 0}(2xycosx
2
−2xy+1)dx+(sinx
2
−x
2
)dy=0
\sf{ \implies \: (2xy \cos {x}^{2}dx + \sin {x}^{2}dy) -( 2xy dx +{x}^{2} dy ) + 1.dx = 0}⟹(2xycosx
2
dx+sinx
2
dy)−(2xydx+x
2
dy)+1.dx=0
\sf{ \implies \: d( y\sin {x}^{2}) -d( {x}^{2} y ) + 1.dx = 0}⟹d(ysinx
2
)−d(x
2
y)+1.dx=0
On integration we get
\displaystyle\sf{ \implies \int d( y\sin {x}^{2}) - \int \: d( {x}^{2} y ) + \int 1.dx = 0}⟹∫d(ysinx
2
)−∫d(x
2
y)+∫1.dx=0
\displaystyle\sf{ \implies y\sin {x}^{2} - {x}^{2} y + x = c}⟹ysinx
2
−x
2
y+x=c
Where C is integration constant
FINAL ANSWER
Hence the required solution is
\sf{ y\sin {x}^{2} - {x}^{2} y + x = c}ysinx
2
−x
2
y+x=c