# Show that the first order differential equation (x + 1)y'-3y= (x+1)^5 is of the linear type. Hence solve for y given that y = 1.5 when x = 0

Question
Show that the first order differential equation $$(x + 1)y'-3y= (x+1)^5$$ is of the linear type.
Hence solve for y given that y = 1.5 when x = 0

2021-01-16
$$(x+1)y'−3y=(x+1)^5$$
$$y'−3/((x+1))y=(x+1)^4$$
Compare with
$$y'+P(x)y= Q(x)$$
$$P(x)= -3/(x+1)$$
$$Q(x)= (x+1)^4$$
Hence given differential equation is of the linear type
$$I.F.= e^(int P(x)dx)$$
$$= e^(int -3/(x+1)dx)\( \(= e^(-3 ln(x+1))$$
$$= e^(ln (x+1)^-3)$$
$$= (x+1)^-3$$
$$y*I.F.= int Q(x)*I.F. dx$$
$$y(x+1)^3= int (x+1)^4*(x+1)^-3 dx$$
$$y(x+1)^-3= int (x+1)dx$$
$$y(x+1)^-3= x^2/2+x+C$$
$$y=(x^2/2+x+C)(x+1)^3$$
$$y(0)=1.5$$
$$(0+0+C)(0+1)^3=1.5$$
C=1.5
$$y=(x^2/2+x+1.5)(x+1)^3$$

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