Find the values of the parameter r for which the function y = e^rx is a solution

Question

Find the values of the parameter r for which the function y = e^rx is a solution of the equation y" - 11y'+ 30y = 0. Find the values of the parameter r for which the function y = e^x sin (rx) is a solution of the equation y" - 2y' + 50y = 0. Find the general solution of the equation y' = 2e^2x/e^4x + 1. Find the solution of the equation y' = 2xe^x^2, satisfying the condition y(0) = 2. Find the general solution of the equation y'= (1 + e^y)e^x - y. Find the solution of the equation y' + 2xy^2 = 0, satisfying the condition y(1) = 1. Find the general solution of the equation (1 + e^x^2) y'+ 3x^2e^x^2y = e^x. Find the general solution of the equation (2 + sin (x))y' + cos (x) y = 2x. Find the solution of the equation y' = x^2 + 2y^2/2xy, satisfying the condition y(1) = 0. Let y be the solution of the equation y' = 2 cos (pi xy) - 2, satisfying the condition y(0) = 1. Apply Euler's method with the horizontal step size h = 1/2 to find an approximation of y(1). Find the general solution of the equation y' = 2 + 3x + 2y + 3xy. Let y be the solution of the equation y' = 4x^2 - y^3

Explanation / Answer

2) we have given y=exsin(rx) and equation is y''-2y'+50y=0

y'=rexcos(rx)+sin(rx)ex

y''=r[ex(-r)sin(rx)+cos(rx)ex]+sin(rx)ex+excos(rx)(r) =-r2sin(rx)ex+2rexcos(rx)+sin(rx)ex

substitute above values into y''-2y'+50y=0

-r2sin(rx)ex+2rexcos(rx)+sin(rx)ex-2(rexcos(rx)+sin(rx)ex)+50exsin(rx)=0

exsin(rx)(-r2-1+50)=0

-r2-1+50=0

r2 =49 implies r=7,-7

solutions are y=exsin(7x) and y=exsin(-7x)

3) we have given y'=(2e2x)/(e4x+1)

dy=(2e2x)/(e4x+1)dx

applying integration

integration of dy =integration of (2e2x)/(e4x+1)dx ---(1)

we solve for integration of (2e2x)/(e4x+1)dx

Substitute u=e2x

du/dx =2e2x implies du=2e2xdx

integration of (2e2x)/(e4x+1)dx =integration of du/(1+u2) =arctan(u)+C=arctan(e2x )+C since resubstitute u=e2x

integration of dy =integration of (2e2x)/(e4x+1)dx

y=arctan(e2x )+C

4) we have given y'=2xe^(x^2)

dy/dx =2xe^(x^2)

dy=2xe^(x^2)dx

applying integration both sides

y=e^(x^2)+C

2=e^(0^2)+C implies C=1 since y(0)=2

y=e^(x^2)+1

since integration of 2xe^(x^2)dx

substitute u=x^2 du=2xdx

integration of 2xe^(x^2)dx =integration of e^t du =e^t +C =e^(x^2) +C

5) we have given y'=(1+ey)ex-y

dy/dx=((1+ey)ex)/ey

ey/(1+ey) dy =exdx

apply integration both sides

integration of ey/(1+ey) dy =integration of exdx

ln(1+ey) =ex+C

(1+ey) =e(e^x +C)

ey =e(e^x +C) -1=e(e^x +C) -e^0=e(e^x +C)/e^0

y=(e^x +C)

6) we have given y'=-2xy2 and y(1)=1

dy/dx =-2xy2

dy/y2 =-2xdx

apply integration both sides

y-1/(-1) =(-2x2)/2+C

-1/y =-x2+C

y=-1/(-x2+C) since by cross multiplication

1=-1/(-(1)2+C) since y(1)=1

-1+C=-1 implies C=0

y=-1/(-x2+0) =1/x2

y=1/x2

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