Substitutions: Question 2-5.2 & Solution

Question 2

On $x>0$ , consider $y'=\frac yx+(y/x-1)(y/x-2), \qquad y(1)=3.$

Tasks

Use $v=y/x$ to derive a separable equation. Find every constant value of $v$ before dividing by a factor involving $v$ .
Solve the transformed equation, and give the resulting original solution family together with any missing straight-line solution.
Select the IVP solution and its maximal interval containing $1$ .
Verify the IVP solution and explain why it cannot cross either straight-line solution within that interval.

Original worksheet page 1: question and worked solution for 2-5-002

Show solutionHide solution+

Question 2 – Solution

Strategy. Constant ratios correspond to straight lines, so preserve them before separating the nonconstant ratio equation.

Step 1: Transform and preserve the constants. Writing $y=xv$ gives $xv'=(v-1)(v-2)$ . The constant ratios $v=1,2$ yield the solutions $\boxed{y=x,\ y=2x}$ . For other ratios, $\int\frac{dv}{(v-1)(v-2)}=\int\frac{dx}{x},\qquad \ln\left|\frac{v-2}{v-1}\right|=\ln x+C_0.$ Thus $(v-2)/(v-1)=Cx$ , with nonzero $C$ for nonconstant $v$ , and $y=\frac{x(2-Cx)}{1-Cx}.$ Allowing $C=0$ restores $y=2x$ ; $y=x$ remains a separate solution. Each formula is used only on intervals where its denominator is nonzero.

Step 2: Select the IVP branch. The condition $v(1)=3$ gives $C=1/2$ , hence $\boxed{y=\frac{x(2-x/2)}{1-x/2},\qquad I=(0,2)}.$ The coefficient domain excludes $0$ , while $y\to+\infty$ as $x\uparrow 2$ .

Step 3: Verify and compare. For $v=(2-Cx)/(1-Cx)$ , $v'=C/(1-Cx)^2$ , and $(v-1)(v-2)=Cx/(1-Cx)^2=xv'$ . Thus $y'=v+xv'$ recovers the original equation, and the selected initial value is $3$ .

On $0<x<2$ , $v=1+1/(1-x/2)>2$ . Therefore $y>2x>x$ throughout the interval: neither line is crossed.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 2-5-002

Substitutions — Question 2

Question 2

Question 2 – Solution