I'm desparately trying to solve this - yes, it's for a homework assignment, I'd much rather someone give me a hint, instead of solving it for me -

y''=-6exp(y)

I'm trying to solve it by setting dy/dx = v, and dv/dx=(dv/dy)*(dy/dx)=v*dv/dx, so v*v'=-6exp(y) (which is an already separated first order ODE).

(v^2)/2=-6exp(y)+c1 ; since v is dy/dx, you can set the right side equal to v(multiply by 2 and take the square root), divide by the right to separate again, and integrate with respect to y

v^2=-12exp(y)+c1 (2*c=a different constant, but still constant).

v=+- (c1-12exp(y))^(1/2)

dy/dx=+- (c1-12exp(y))^(1/2)

int(((c1-12(exp(y)))^-(1/2))dy)=int(dx)

solve the left with a trigonometric substitution, 12exp(y)=c*cos^2(t)

I won't show all of my work at this juncture, but I'm running into problems after the integration, and just want some advice (am I doing this right? On the right track? Hopelessly lost?).

y''=-6exp(y)

I'm trying to solve it by setting dy/dx = v, and dv/dx=(dv/dy)*(dy/dx)=v*dv/dx, so v*v'=-6exp(y) (which is an already separated first order ODE).

(v^2)/2=-6exp(y)+c1 ; since v is dy/dx, you can set the right side equal to v(multiply by 2 and take the square root), divide by the right to separate again, and integrate with respect to y

v^2=-12exp(y)+c1 (2*c=a different constant, but still constant).

v=+- (c1-12exp(y))^(1/2)

dy/dx=+- (c1-12exp(y))^(1/2)

int(((c1-12(exp(y)))^-(1/2))dy)=int(dx)

solve the left with a trigonometric substitution, 12exp(y)=c*cos^2(t)

I won't show all of my work at this juncture, but I'm running into problems after the integration, and just want some advice (am I doing this right? On the right track? Hopelessly lost?).

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