Answer to Question #197054 in Differential Equations for Rabia

Question

Show that the functions defined below is solution of the Differential Equations or not

a)𝑦=𝑥+3𝑒−𝑥, 𝑦′+𝑦=𝑥+1,𝑤h𝑒𝑟𝑒 𝑦′=𝑑𝑦 𝑑𝑥

b)𝑦=2𝑥2 +𝑒𝑥 +6𝑥+7, 𝑦′′ −3𝑦′ +2𝑦=4𝑥2 c)𝑦= 1 , (1+𝑥2)𝑦′′+4𝑥𝑦′+2𝑦=0

d) 𝑢 = 𝑒3𝑥 cos 2𝑦 , 𝑢𝑥𝑥 − 2 𝑢𝑦𝑦 = 17 𝑢 , 𝑢𝑥𝑥 = 𝑑2𝑢 𝑑𝑥2

Answer 1

(a) $y=x+3e^{-x}$

$\Rightarrow y'=1-3e^{-x}$

Taking LHS-

$y'+y=1-3^{-x}+1+3^{-x}=2\neq$ RHS

Given y is not the solution.

(b) $y=2x^2+e^x+6x+7$

$\Rightarrow y'=4x+e^x+6\\[9pt]\Rightarrow y''=4+e^x$

Taking LHS-

$y''-3y'+2y\\=4+e^x-12x-3e^x-18+2x^2+e^x+6x+7\\=-e^x+2x^2-6x+11\neq RHS$

Given y is not the solution.

(c)𝑦= 1 , $(1+𝑥^2)𝑦''+4𝑥𝑦'+2𝑦=0$

As, y=1

$\Rightarrow y'=0,y''=0$

Taking LHS-

$(1+𝑥^2)𝑦′′+4𝑥𝑦′+2𝑦\\=(1+x^2)0+4x(0)+2(1)\\=2\neq RHS$

Hence y=1 is not the solution.

(d)

$𝑢 = 𝑒^{3𝑥} cos 2𝑦\\[9pt]\Rightarrow u_x=3e^{3x}cos2y\\[9pt]\Rightarrow u_{xx}=9e^{3x}cos2y$

$\Rightarrow u_y=-2e^{3x}sin2y\\[9pt]\Rightarrow u_{yy}=-4e^{3x}cos2y$

Taking LHS-

$𝑢_{𝑥𝑥} − 2 𝑢_{𝑦𝑦}\\=9e^{3x}cos2y+8e^{3x}cos2y\\=17e^{3x}cos2y=17u=RHS$

Given y is the solution.