Question

Question 1 5 pts What is the order of the DE? Is the DE linear or nonlinear? 3 54' - (tan x)y= Vx+ +1 (7) 5.22 3rd order; nonlinear 2nd order; nonlinear 2nd order; linear 3rd order; linear

Answer 1

5724" + tan(2y = V.x2 +1(y) 3

This is a non-linear second order differential equation.

2nd Order; nonlinear.

2)

$y=2e^{-3x}-4e^{5x}$

The derivatives of y are

$y'=-6e^{-3x}-20e^{5x}$

and

$y''=18e^{-3x}-100e^{5x}$

Now,

-Зr 100e5х + 12e — Зr 5r у" - 2y - 15y = 18e + 40e5 -Зr – 30e + 60e5г

$y''-2y'-15y=30e^{-3x}-100e^{5x}+100e^{5x}-30e^{-3x}=0$

y" – 2y – 15y = 0

Hence y is the solution of the given differential equation.

3)

dy dc = 3.xy + 2y – 15.6 – 10

$\implies \dfrac{dy}{dx}=y(3x+2)-5(3x+2)$

dy dc (3.1 +2)(y-5)

$\implies \dfrac{dy}{(y-5)}=(3x+2)dx$

Integrate on both sides, we have

$\int \dfrac{dy}{(y-5)}=\int (3x+2)dx$

$\implies \ln|(y-5)|=(3/2)x^2+2x+\ln C$

$\implies y-5=e^{(3/2)x^2+2x+\ln C}$

$\implies y=\bf5+Ce^{(3/2)x^2+2x}$

4)

(-3, 0)

5)

$y'+(2\sin(2x)y=\sin(2x)$

$\implies dy/dx=-(2\sin(2x)y+\sin(2x)=\sin(2x)*(1-2y)$

$\implies \dfrac{dy}{y-1/2}=(-2)\sin(2x)dx$

Integrate on both sides, we have

$\implies \int\dfrac{dy}{y-1/2}=(-2)\int\sin(2x)dx$

$\implies \ln|y-1/2|=\cos(2x)+\ln C$

$\implies y-1/2=e^{\cos(2x)+\ln C}$

$\implies y=\bf \dfrac{1}{2}+Ce^{\cos(2x)}$

6)

$(1+3x^2y^4)dx=-4x^3y^3dy$

$(1+3x^2y^4)dx+4x^3y^3dy=0$

Comparing with M dx+N dy=0, we have

$M=(1+3x^2y^4),\qquad N=4x^3y^3$

The equation is exact and the solution is given by

$\int (1+3x^2y^4)dx+\int 0dy=C$

Leave those terms in N which contains x, because they already comes in M.

$x+x^3y^4=C$

$\bf x+x^3y^4=C$

Please comment if you have any doubt or answer is mismatch.

Thank you......