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

Answers (1)

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
\(y=(x^2/2+x+1.5)(x+1)^3\)
0

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