Question

# If xy+6e^{y} = 6e, find the value of y" at the point where x = 0.

Second order linear equations
If $$\displaystyle{x}{y}+{6}{e}^{{{y}}}={6}{e}$$, find the value of y" at the point where x = 0.
Here the Differential equations is given by $$xy+6e^y=6e$$ with y(0)=1y(0)=1. Taking differentiate both sides with respect to x we get $$(d/dx(xy+6e^y))=(d/dx)6e^y (d/dx(xy)+6(d/dx)e^y=0 y+xy'+(6e^y)y'=0$$ From here putting y(0)=1y(0)=1 we get y′(0)=−1/6e. Agin differentiate both sides above expressions with respect to x we get $$(d/dx)(y+xy'+(6e^y)y')=0 y'+y'+xy''+(6e^y)y'^2+(6e^y)y''=0$$ Now putting x=0x=0 and y(0)=1y(0)=1, y′(0)=−1/6e we get -$$(1/3e)-6e(1/36e^2)+6ey''=0 -> 6ey''=1/3e+1/6e -> 6ey''=1/2e -> y''=1/12e^2$$