Solve the initial-value problem?

Question

dy/dx = 2/5.sin(2x).e^cos x .y^2 (y>0)

y = 1 when x = 0.

giving the solution in explicit form.

Answer 1

The equation is separable, and can be written in the form
dy/y^(-2) = (2/5)*(sin (2x))*(e^(cos (x))). By using the double angle formula for sin (2x), we can write this as
dy/y^(-2) = (4/5)*(sin (x))*(cos (x))*e^(cos (x)). The left side integrates as -y^(-1). For the left side, first make the substitution w = cos (x), then integrate by parts with u = w and dv = (e^u)du. We find -y^(-1) = (4/5)*[-(cos (x))*e^(cos (x)) + e^(cos (x))] + C. Using y = 1 when x = 0, we finf C = 1.

Answer 2

I'll try.... Hope I get this right...

1 = y(4/5)(e^cos x)(cos x - 1) + y...

hmmm...

well first, group the terms with the x and the y together...

dy/y^2 = 2/5(sin 2x)(e^cos x)(dx)

then integrate...

-1/y = -4/5 (e^cos x)(cos x - 1) + C (C is the constant...)

substitute x = 0 and y = 1...

1 = (4/5) (e^cos 0)(cos 0 - 1) + C (I've cancelled the -1 in both sides...

we know that...

cos 0 = 1...

so, we are left with C = 1...

substituting and manipulating the whole equation, we will get the answer I gave above...

Answer 3

This can be solved by separating the variables.

∫y-2 dy = ∫2/5.sin(2x)e^cosx

-1/y =2/5∫2sinxcosxe^cosx

Then take it from there. You can integrate the RHS by parts and then use the Reduction Formulae technique because it is "cyclic".

Answer 4

your question is wrong with dec point however below is a corrected sulution...but prob not right as its only guess as to what you meant..

nsolve(d y/d x=2/5 sin(2 x) e^cos(x y^2) (y>0), {x})

Answer 5

my son would solve this.