Solve the following differential equation:&nbsp;1+y'(x)2-y''(x)(y(x)-λ)=0

Let u=y−λ, v=u′=y′. then we can rewrite the differential equation as1+v2−v′u=0→udvdx=v′u=1+v2, dudx=u′=vwhich impliesudvdu=1+v2v→2vdv1+v2=2duuon integration, we get(1+v2)=cu2→v=±1−cu2→du1−cu2=±dxnow the integration will depend on the sign of c: if c&gt;0, you get an inverse sine, sin−1 and if c&lt;0, then ln⁡

Find out the answer to "Solve the following differential equation:1+y'(x)2-y''(x)(y(x)-λ)=0"

Schwelliney

Answered question

2021-11-23

Solve the following differential equation: $1 + y' (x)^{2} - y'' (x) (y (x) - λ) = 0$

Answer & Explanation

Helen Rodriguez

Beginner2021-11-24Added 9 answers

Let $u = y - λ, v = u^{'} = y^{'}$ . then we can rewrite the differential equation as
$1 + v^{2} - v^{'} u = 0 \to u \frac{d v}{d x} = v^{'} u = 1 + v^{2}, \frac{d u}{d x} = u^{'} = v$
which implies
$u \frac{d v}{d u} = \frac{1 + v^{2}}{v} \to \frac{2 v d v}{1 + v^{2}} = \frac{2 d u}{u}$
on integration, we get
$(1 + v^{2}) = c u^{2} \to v = \pm \sqrt{1 - c u^{2}} \to \frac{d u}{\sqrt{1 - c u^{2}}} = \pm d x$
now the integration will depend on the sign of c: if $c > 0$ , you get an inverse sine, $\sin^{- 1}$ and if $c < 0$ , then $\ln$

Forneadil

Beginner2021-11-25Added 18 answers

As abel noted, with

u = y - λ

we get

1 + {(u^{'})}^{2} - u^{″} u = 0

. Now if you recall that

{cosh}^{2} - {sinh}^{2} = 1

and

cosh^{″} = cosh

, you are led to consider something of the form

u = a cosh (b x + c)

and we find that this works if

b = \frac{1}{a}

. Thus we get the solution

y = λ + a cosh (\frac{x}{a} + c)

where

a \neq 0

and c are arbitrary parameters.

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