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Question

Mathematics Question on Differential equations

The equation of the curve passing through the point (1, 1) such that the slope of the tangent at any point (x, y) is equal to the product of its co-ordinates is

A

2log⁡y=x2+12 \log y = x^2 + 1

B

2log⁡x=y2−1 2 \log x = y^2 - 1

C

2log⁡x=y2+12 \log x =y^2 + 1

D

2log⁡y=x2−12 \log y = x^2 - 1

Answer

2log⁡y=x2−12 \log y = x^2 - 1

Explanation

Solution

dydx=xy\frac{d y}{d x}=xy
1ydy=xdx\frac{1}{y} dy=xdx
IntegrateIntegrate
logy=x22+c…(1)log y=\frac{x^{2}}{2}+c\quad\ldots\left(1\right)
Equation(1)passing through(1,1)Equation \left(1\right) passing \,through \left(1, 1\right)
log(1)=122+Clog\left(1\right)=\frac{1^{2}}{2}+C
0=12+c0=\frac{1}{2}+c
c=−12c=-\frac{1}{2}
log y=x22−12log\, y=\frac{x^{2}}{2}-\frac{1}{2}
2 log y=x2−12 \,log \, y=x^{2}-1