Answer to Question #272661 in Differential Equations for Lando

Question

With the use of reduction of order for differential

equations, reduce the following to first order and

solve.

(i) 𝑦

′′ + 𝑒

𝑦𝑦

′3 = 0,

(ii) 𝑥𝑦

′′ + 2𝑦

′ + 𝑥𝑦 = 0, 𝑦1 =

sin 𝑥

𝑥

,

(iii) (1 − 𝑥

2

)𝑦

′′ − 2𝑥𝑦

′ + 2𝑦 = 0, 𝑦1 = 𝑥,

(iv) 4𝑥

2𝑦

′′ − 3𝑦 = 0, 𝑦(1) = 3, 𝑦

′

(1) = 2.5

Answer 1

i)

$𝑦'' + 𝑒^𝑦(𝑦')^3 = 0$

$y'=u$

$y''=uu'$

$uu'+e^yu^3=0$

$u'+e^yu^2=0$

$du/u^2=-e^ydy$

$1/u=e^y+c$

$\frac{}{}$ $dx=(e^y+c)dy$

$x=e^y+cy+c_1$

ii)

$y''+p(x)y'+q(x)y=0$

$y''+2y'/x+y=0$

$y_2=v(x)y_1=vsinx/x$

$v(x)=c\int\frac{1}{y_1^2}e^{-\int p(x)dx}dx=c\int\frac{x^2}{sin^2x}e^{-2lnx}dx=$

$=c\int\frac{1}{sin^2x}dx=c(-cotx)+k$

$v(x)=cotx$

$y_2=cotxsinx/x=cosx/x$

$y(x)=c_1sinx/x+c_2cosx/x$

iii)

$𝑦 ''− 2𝑥𝑦 '/(1 − 𝑥 ^2 )+ 2𝑦/(1 − 𝑥^ 2 ) = 0$

$v(x)=c\int\frac{1}{y_1^2}e^{-\int p(x)dx}dx=c\int\frac{1}{x^2}e^{-ln(x^2-1)}dx=c\int\frac{1}{x^2(x^2-1)}dx=$

$=c(\frac{ln\frac{x-1}{x+1}}{2}+\frac{1}{x})+k$

$v(x)=\frac{ln\frac{x-1}{x+1}}{2}+\frac{1}{x}$

$y_2=x(\frac{ln\frac{x-1}{x+1}}{2}+\frac{1}{x})=\frac{xln\frac{x-1}{x+1}}{2}+1$

$y(x)=c_1x+c_2(\frac{xln\frac{x-1}{x+1}}{2}+1)$

iv)

$y=x^k$

$(x^k)''=k(k-1)x^{k-2}$

$(4k^2-4k-3)x^k=0$

$4k^2-4k-3=0$

$k_1=-1/2,k_2=3/2$

$y_1=c_1/\sqrt x,y_2=c_2x^{3/2}$

$y(x)=y_1+y_2=c_1/\sqrt x+c_2x^{3/2}$

$y(1)=c_1+c_2=3$

$y'(x)=-c_1/(2x\sqrt x)+3c_2x^{1/2}/2$

$𝑦 '(1) =-c_1/2+3c_2/2= 2.5$

$c_2-3+3c_2=5$

$c_2=2,c_1=3-c_2=1$

$y(x)=1/\sqrt x+2x^{3/2}$