Find the equation of a curve that passes through (10, 1) and whose slope at any point is equal to

Slope of the tangent to the curve at (x,y) = dy/dx

Given equation:

dy/dx = x+ xy

dy/dx - xy = x

Differential equation is of the form

dy/dx + PY = Q

where  P = -x and Q = x

\(IF\ =\ e^{f\ pdx}\ \)

\(=\ e^{-f\ xdx}\ =\ e^{\frac{-x^2}{2}}\)

Solution is

\(y\ \left(IF\right)\ =\ f\left(Q\ x\ IF\right)dx\ +C\)

\(y\ e^{\frac{-x^2}{2}=\ f\ xe^{\frac{-x^2}{2}}dx\ +C}\)

putting -x2/2 = t

-2x/2 dx = dt

x dx = -dt

Thus, our equations become

\(ye^{\frac{-x^2}{2}}=f-e^tdt\ +C\)

\(ye^{\frac{-x^2}{2}}=\ -e^t+C\)

putting t = -x2/2

\(ye^{\frac{-x^2}{2}}=-e^{\frac{-x^2}{2}}+C\)

\(\frac{y}{e^{\frac{-x^2}{2}}}=\ -\ \frac{1}{e^{\frac{x^2}{2}}}+C\)

\(y=-1\ +\ Ce^{\frac{x^2}{2}}....\left(1\right)\)

Since curve passes through (0,1)

putting x=0, y=1 in (1)

\(y\ =\ -1\ +Ce^{\frac{x^2}{2}}\)

\(1\ =\ -1\ +Ce^{\frac{0^2}{2}}\)

1 = -1 +C
1+1=C
C=2

putting value of C in (1)

\(y=-1+Ce^{\frac{x^2}{2}}\)

\(y=\ -1\ +\ 2e^{\frac{x^2}{2}}\)

therefore, the equation of the curve is \(y=\ -1\ +\ 2e^{\frac{x^2}{2}}\)

Nice thanks