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Findthe equation of a curve passing through the origin given that the slope ofthe tangent to the curve at any point (x, y) is equal to the sum of thecoordinates of the point. |
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Answer» Let the moving point be `(x,y)` Given that, `(dy)/(dx)=x+yimplies (dy)/(dx)-y=x` Here, `P=-1`, `Q=x` `:. I.F.=e^(int-1dx)=e^(-x)` and general solution : `y(e^(-x))=intxe^(-x)dx+c` `implies ye^(-x)=-xe^(-x)-int1(-e^(-x))dx+c` `=-xe^(-x)-e^(-x)+c` `implies y=-x-1+ce^(x)` This curve passes through `(0,0)` `:. 0=0-1+cimpliesc=1` `:.`Curve is `y=-x-1+e^(x)` `implies x+y+1=e^(x)` |
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