Write the equation in the form y\' = f(y/x) then use the substitution y = xu to

Question

Write the equation in the form y' = f(y/x) then use the substitution y = xu to find an implicit general solution. Then solve the initial value problem. y' = 4y^2 + 3x^2/xy, y(1) = 1 The resulting differential equation in x and u can be written as xu' = which is separable. Separating variables we arrive at du = dx/x Separating variables and simplifying the solution can be written in the form u^2 + 1 = Cf(x) where C is an arbitrary constant and f(x) = Transforming back into the variables x and y and using the initial condition to find C we find the explicit solution of the initial value problem is y =

Explanation / Answer

We have given y'=(4y2+3x2)/xy and y(1)=1

y'=4(y/x)+3(x/y)

substitute y=xu,y'=xu'+u

xu'+u=4u+3/u

xu'=3u+3/u

xu'=3(u^2+1)/u

by cross multiplication

x*u*du=3(u^2+1)dx

again by cross multiplication

(u/3(u^2+1))du =dx/x

integrating both sides

integration of (u/3(u^2+1))du =integration of (dx/x)

substitute v=u^2+1,dv=2udu implies udu=dv/2

(1/3)*integration of (1/(v))(dv/2) =ln(x)

(1/6)*(ln(v))=ln(x)

resubstitute v=u^2+1

(1/6)*(ln(u^2+1))=ln(x)

ln(u^2+1)=6*ln(x)+C

u^2+1=e^(6*ln(x)+C)=e^(6*ln(x))*e^C

u^2+1=e^(6*ln(x))*C since e^C=C

by comparing with u^2+1=f(x)*C we get f(x)=e^(6*ln(x))=x^6

u^2+1=(e^(ln(x)))^6*C since e^(px)=(e^x)^p

u^2+1=(x)^6*C

x^6=(u^2+1)/C

substitute u=y/x

x^6=((y/x)^2+1)/C

substitute y(1)=1

1=((1/1)^2+1)/C

1=2/C implies C=2

x^6=((y/x)^2+1)/2 since C=2

y^2/x^2=2x^6-1

y^2=2x^8-x^2

y=sqrt(2x^8-x^2)

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