Find an equation of the form y=f(x) for a curve with x-intercept 1 whose tangent line at any point (x,y) has slope x^(3)e^(-y).

Question

Find an equation of the form y=f(x) for a curve with x-intercept 1 whose tangent line at any point (x,y) has slope x^(3)e^(-y). 

I have done dy/e^y=x^3dx then solving for y = ln(x^4/4)+C but am not sure what to do with the f(1)=0. When I plug in 1 I get ln(1/4)+C but that is not the answer.

Does anyone know where I am going wrong?

Answer 1

dy/dx = x³e^-y; ∫e^ydy = ∫x³dx; e^y = x⁴/4 + C; y = f(x) = ln(x⁴/4 + C); x = 1 y = 0; ln(1/4 + C ) = 0; 1/4 + C = 1;

C = 3/4. So, f(x) = ln(x⁴ + 3) - ln4