solve.(1+x^2)dy = (4x^2-2xy)dx
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Answer: (1+x^2)dy/dx+2xy-4x^2=0
dy/dx+2xy/1+x^2=4x^2/1+x^2
where,y(IF)= integration of Q IF dx+c
y(1+x^2)=integration of 4x^2/1+x^2 (1+x^2)dx+c
y(1+x^2)=4 integration of x^2dx+c=4x^3/3+c
where, y=0,x=0
0(1+0)=4(0/3)+c=c=0
y(1+x^2)=4x^3/3
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