Question

Find the equation of a curve passing through the point \((0,2)\) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer 1

Let \(F(x,y)\) be the curve and let \((x,y)\) be a point on the curve.The slope of the tangent to the curve at \((x,y)\) is \(\frac{dy}{dx}\).
According to the given information:
\(\frac{dy}{dx}+5=x+y\)
\(⇒\frac{dy}{dx}-y=x-5\)
This is a linear differential equation of the form:
\(\frac{dy}{dx}+py=Q\)(where p=-1 and Q=-5)
Now,\(I.F.=_e∫p\,dx=_e∫(-1)dx=e^{-x}.\)
The general equation of the curve is given by the relation,
\(y(I.F.)=∫(Q\times I.F.)dx+C\)
\(⇒y.e^{-x}=∫(x-5)e^{-x}dx+C...(1)\)
Now,\(∫(x-5)e^{-x}dx=(x-5)∫e^{-x}dx-∫[\frac{d}{dx}(x-5).∫e^{-x}dx]dx.\)
\(=(x-5)(-e^{-x})-∫(-e^{-x})dx\)
\(=(5-x)e^{-x}+(-e^{-x})\)
\((4-x)e^{-x}\)
Therefore,equation(1)becomes:
\(y\,e^{-x}=(4-x)e^{-x}+C\)
\(⇒y=4-x+Ce^x\)
\(⇒x+y-4=Ce^x...(2)\)
The curve passes through point(0,2).
Therefore,equatiuon(2)becomes:
\(0+2-4=Ceº\)
\(⇒-2=C\)
\(⇒C=-2\)
Substituting C=-2 in equation(2),we get:
\(x+y-4=-2e^x\)
\(⇒y=4-x-2e^x\)
This is the required equation of the curve.