# Show that y=c1+c2t1/2y = c_1+c_2t^{1/2} won’t always solve yy″+(y′)2=0yy”+(y’)^2=0

Show that y = c_1+c_2t^{1/2}y = c_1+c_2t^{1/2} doesn’t always solve the differential equation

yy”+(y’)^2=0yy”+(y’)^2=0

that’s what I did:

yy”+(y’)^2= \left(c_1+\frac{c_2}{2}t^{-1/2}\right)\left(-\frac{c_2}{4}t^{-3/2}\right)+\left( \frac{c_2}{2}t^{-1/2}\right)^2 = yy”+(y’)^2= \left(c_1+\frac{c_2}{2}t^{-1/2}\right)\left(-\frac{c_2}{4}t^{-3/2}\right)+\left( \frac{c_2}{2}t^{-1/2}\right)^2 =

-c_1\frac{c_2}{4}t^{-3/2}-\frac{c_2^2}{8}t^{-2}+\frac{c_2^2}{4}t^{-1}-c_1\frac{c_2}{4}t^{-3/2}-\frac{c_2^2}{8}t^{-2}+\frac{c_2^2}{4}t^{-1}

but I can t even group the therm to see when they’ll sum to 00. What should I do?

Also, why this doesn’t contradict the theorem that says that if y_1y_1 and y_2y_2 are solutions of a diferential equation, then c_1y_1+c_2y_2c_1y_1+c_2y_2 is also a solution?

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Because superposition principle only holds for linear ode.
– Jacky Chong
Oct 21 at 1:24

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You almost got the answer: y y” + (y’)^2y y” + (y’)^2 is an expression in tt that does not simplify to 00 (except your expression is not quite right), so this is not a solution to the differential equation. What you should have obtained was -c_1 c_2/(4 t^{3/2})-c_1 c_2/(4 t^{3/2}).

BTW, it turns out that the general solution of this one is
y = \pm \sqrt{c_1 + c_2 t} y = \pm \sqrt{c_1 + c_2 t}

when you set

\frac{y^\prime}{y} =-\frac{y^{\prime \prime}}{y\prime} =c_1 \frac{y^\prime}{y} =-\frac{y^{\prime \prime}}{y\prime} =c_1

it suggests chain rule of a product

{y^\prime}{y} = c_1{y^\prime}{y} = c_1

which integrates to

y^2 = c_2+ 2 c_1\,xy^2 = c_2+ 2 c_1\,x

Taking square root for both terms at normal speed you get

y = \sqrt{c_2+ 2 c_1\,x} y = \sqrt{c_2+ 2 c_1\,x}

but at higher speeds the first term gets excluded 🙂

y = c_2+ \sqrt{c_3\,x} y = c_2+ \sqrt{c_3\,x}

It is a parabola with a constant subnormal length c_1.c_1.

Since Jacky answered the bolded question [in the comment on the OP], I’ll answer the main one: you are basically done.

If c_1+c_2t^{1/2}c_1+c_2t^{1/2} always solved the DE, it would in particular have to solve it when t=1t=1 for every c_1c_1 and c_2c_2. But when t=1t=1, then yy”+(y’)^2=(1/8)(-2c_1c_2-c_2^2+2c_2^2)yy”+(y’)^2=(1/8)(-2c_1c_2-c_2^2+2c_2^2). This is not zero for every choice of c_1c_1 and c_2c_2; for instance, if c_1=c_2=1c_1=c_2=1 it is -1/8-1/8.

If y=a+b \sqrt{t}\implies y’=\frac{b}{2 \sqrt{t}}\implies y”=-\frac{b}{4 t^{3/2}}y=a+b \sqrt{t}\implies y’=\frac{b}{2 \sqrt{t}}\implies y”=-\frac{b}{4 t^{3/2}} y y” +(y’)^2=-\frac{a b}{4 t^{3/2}}y y” +(y’)^2=-\frac{a b}{4 t^{3/2}} So, if the rhs has to be 00 either (a=0, b\neq 0)(a=0, b\neq 0) or (b=0, a\neq 0)(b=0, a\neq 0) or (a=0,b=0)(a=0,b=0).