Question

Pers.. dif..linear
mohon bantuannya kk :v

Answer 1

Y' = (x²+3y)/2x
y' =  x/2 + 3y/2x
y' + (-3/2x).y = x/2........faktor integral -3/2x→ e^∫ -3/2x dx
                                                                        e^(-3/2).ln x
                                                                        x^(-3/2)

y.x^(-3/2) = ∫ (x/2).x^(-3/2) 
y.x^(-3/2) = ∫ (1/2√x)
y.x^(-3/2) = x^(1/2) + c1
             y = x^(1/2)/x^(-3/2) + c1/x^(-3/2)
             y = x² + c1.x^(3/2)

Answer 2

Inget,

dy/dx + P(x)*y = Q(x)

dy/dx = (x^2 + 3y)/(2x)
dy/dx = x/2 + (3y)/(2x)
dy/dx - (3/2)(y/x) = x/2
dy/dx - (3/2)y.x^(-1) = x/2
dy/dx + (-(3/2)y.x^(-1)) = x/2

faktor integrasi :
e^∫P(x) dx = e^∫-(3/2)x^(-1) dx
= e^(-3/2.ln(x))
= x^(-3/2)

d/dx.x^(-3/2) + (-3/2)y.x^(-1).x^(-3/2) = x^(3/2).x/2

inget,
y = u.v
y' = u'v + uv'

maka,
u = y
v = x^(-3/2)

sehingga,
y.x^(-3/2) = ∫x^(-3/2).x/2 dx
y.x^(-3/2) = ∫x^(-1/2)/2 dx
y.x^(-3/2) = x^(1/2) + c
y = x^2 + x^(3/2).c