Solve x(dy/dx)=y + cos^2(y/x)

i just got this differential equation out of the blue, and i haven't done these in quite a while. so please help me solve this one.

hey mate,

I forget the name of the substitution method I am going to use, however I will use

v(x) = y(x)/x

or

y(x) = x * v(x)

Thus,

y'(x) = v(x) + x * v'(x)

sub into your DE

x * ( v(x) + x* v'(x) ) = x*v(x) + ( cos(v) )^2

x*v(x) + x^2 * v'(x) = x*v(x) + (cos(v))^2

x^2 * v'(x) = ( cos(v) )^2

v'(x) can be written as dv/dx (...obviously)

so

x^2 * dv/dx = (cos(v))^2

(1/(cos(v))^2) dv = (1/x^2) dx

sec^2(v) dv = (1/x^2) dx

integrate both sides

tan(v) = -1/x + C

v = arctan( -1/x + C) ..................C = integration constant

recall y = x * v(x)

Thus,

y = x * arctan ( -1/x + C)

You may want to double check my algebra...

Hope this helps,

David

solve x(dy/dx)=y + cos^2(y/x)
Sol (dy/dx)=y/x + (1/x)cos^2(y/x) .................(i)
Put y/x = u or y = ux
Differentiate dy/dx = u + x(du/dx)..................(ii)
u + x(du/dx) = u +(1/x) cos^2(u)
x(du/dx) = (1/x) cos^2(u)
du/cos^2(u) = dx/x^2
sec^2(u)*du = dx/x^2
integrate
tan(u) = -1/x +C
tan(y/x) = -1/x +C
y/x = tan^-1{-1/x+C}
y = x tan^-1{-1/x+C} ...................Ans

x dy/dx = y + cos²(y/x)
dy/dx = y/x + cos²(y/x)/x

This is not separable, so we'll use a substitution:
u = y/x
y = ux
dy/dx = u + x du/dx

u + x du/dx = y/x + cos²(y/x)/x
u + x du/dx = u + cos²(u)/x
x du/dx = cos²(u)/x
1/cos²u du = 1/x² dx

Integrate both sides:
∫ 1/cos²u du = ∫ 1/x² dx
∫ sec²u du = ∫ 1/x² dx
tan(u) = -1/x + C
tan(y/x) = -1/x + C
y/x = tan⁻¹(C - 1/x)

y = x tan⁻¹(C - 1/x)

Mαthmφm

xdy/dx=y + cos^2(y/x)
Put y=vx
So, dy/dx=v+xdv/dx
dy/dx=v+cos^2(v)/x
v+xdv/dx=v+cos^2(v)/x
Or, dv/dx=cos^2(v)/x^2
Or, dv/cos^2(v)=dx/x^2
Integrate both sides.
tan(v) = -1/x + C
As, v=y/x
Thus, we get
tan(y/x)=-1/x + C <--- ANSWER