Question

dy/dx - ay = Q, where a is a constant, giving your snswer in terms of of a, when

a) Q = ke^lamda(x)
b) Q = ke^ax
c) Q = (kx^n)(e^ax)

12) use the substitution z = y^-1 to transform the differntial equation x dy/dx +y = (y^2)ln(x), into a linear equation. hense obtain the general solution of the original equation.

13) use the substitution z = y^2 to transform the differential equation 2cosx dy/dx - ysinx +y^-1 = 0, into a linear equation. hense obtain the general solution of the original equation.

14) us the substitution u = x + y to transform the differential equation dy/dx = (x +y +1)(x + y - 1) into a differential equation in u and x. by first solving this new equation. hense obtain the general solution of the original equation

15) use the substitution u = y - x - 2 to transform the differential equation dy/dx = (y - x - 2)^2 into a differntial equation in u and x. By first solving this new equation, find the general solution of the original equation, giving y in therms of x.

Accepted Answer

begin{array}{l} n frac{dy}{dx} - ay = 0   n frac{dy}{y} = adx   n ln y = ax +  ln C   ny = C(x)e^{ax}   n frac{dy}{dx} =  frac{dC}{dx}e^{ax} + Cae^{ax}   n end{array}  a)   begin{array}{l} n frac{dC}{dx}e^{ax} + Cae^{ax} - Cae^{ax} = ke^{ lambda(x)}   n frac{dC}{dx} =  frac{ke^{ lambda(x)}}{e^{ax}}   nC(x) =  int  frac{ke^{ lambda(x)}}{e^{ax}} dx   n end{array}  Answer:  y =  left( int  frac{ke^{ lambda(x)}}{e^{ax}} dx right) * e^{ax}  B)   begin{array}{l} n frac{dC}{dx}e^{ax} + Cae^{ax} - Cae^{ax} = ke^{ax}   n frac{dC}{dx} =  frac{ke^{ax}}{e^{ax}}   nC(x) = k int x ,dx = kx + c_1   n end{array}  Answer:  y = (kx + c_1) * e^{ax}  c)   begin{array}{l} n frac{dC}{dx}e^{ax} + Cae^{ax} - Cae^{ax} = kx^n e^{ax}   n frac{dC}{dx} = kx^n   nC(x) = k int x^n ,dx = k  frac{x^{n+1}}{n+1} + c_1   n end{array}  Answer:  y =  left(k  frac{x^{n+1}}{n+1} + c_1 right) * e^{ax}  12)   frac{x dy}{dx} + y = y^2 l n x z =  frac{1}{y} d z = - y^{-2} d y d y = - y^{2} d z -  frac{x y^{2} d z}{d x} + y = y^{2} l n x -  frac{x d z}{d x} +  frac{1}{y} = l n x -  frac{x d z}{d x} + z = l n x -  frac{d z}{d x} +  frac{z}{x} =  frac{l n x}{x} t =  frac{z}{x} z = t x d z = t d x + x d t  frac{t d x + x d t}{d x} - t =  frac{l n x}{x} t +  frac{x d t}{d x} - t =  frac{l n x}{x} t =  int  frac{l n x}{x^{2}} d x =  left(u = l n x,  , dv =  frac{1}{x^{2}} d x right) = u v -  int v d u = -  frac{1}{x} l n x +  int  frac{1}{x^{2}} = -  frac{1}{x} l n x -  frac{1}{x} + c_{1} z =  left(- frac{1}{x} l n x -  frac{1}{x} + c_{1} right) x  Answer:  y =  frac{1}{z} =  frac{1}{ left(- frac{1}{x} l n x -  frac{1}{x} + c_{1} right) x}  13)  2 c o s x  frac{d y}{d x} - y s i n x +  frac{1}{y} = 0 z = y^{2} d z = 2 y d y  frac{2  cos x}{2 y}  frac{dz}{dx} - y  sin x +  frac{1}{y} = 0  cos x  frac{dz}{dx} - y^{2}  sin x + 1 = 0  cos x  frac{dz}{dx} - z  sin x + 1 = 0  frac{dz}{dx} -  frac{z  sin x}{ cos x} +  frac{1}{ cos x} = 0  frac{dz}{dx} - z  tan x +  frac{1}{ cos x} = 0  frac{dz}{dx} - z  tan x = 0  frac{dz}{dx} = z  tan x  frac{dz}{z} =  tan x  , dx  ln z = - ln( cos x) +  ln c z_{1} = - cos x + c z_{2} = - frac{A}{ cos x}  frac{dz}{dx} = A  frac{ tan x}{ cos x} A  frac{ tan x}{ cos x} -  left(- frac{A}{ cos x} right)  tan x = - frac{1}{ cos x} 2A  frac{ tan x}{ cos x} = - frac{1}{ cos x} 2A  tan x = -1 A = - frac{1}{ tan x} z_{2} =  frac{1}{ tan x}  frac{1}{ cos x}  Answer:  z = z_{1} + z_{2} =  frac{1}{ tan x}  frac{1}{ cos x} -  cos x + c  14)  { frac {dy}{dx}}=(x+y+1)(x+y-1)  u=x+y  du=dx+dy  dy=du-dx  { frac {du-dx}{dx}}=(u+1)(u-1)  { frac {du}{dx}}-1=(u+1)(u-1)  { tfrac {du}{dx}}=u^{2} -1+1  { frac {du}{dx}}=u^{2}  { frac {du}{u^{2}}}=dx   int { frac {du}{u^{2}}}= int dx  x=-{ frac {1}{u}}+c  u={ frac {1}{c-x}}  { frac {1}{c-x}}=x+y  **Answer:**  y=x-{ frac {1}{c-x}}  15)  { frac {dy}{dx}}=(y-x-2)^{2}  u=y-x-2  du=dy-dx  dy=du+dx  { frac {du+dx}{dx}}=(u)^{2}  { frac {du}{dx}}=(u)^{2}-1   int  frac {du}{(u)^2 - 1} =  int dx x =  frac{1}{2}  left(  frac{u - 1}{u + 1}  right) + c u =  frac{1 + e^{2(x + c)}}{1 - e^{2(x + c)}} y - x - 2 =  frac{1 + e^{2(x + c)}}{1 - e^{2(x + c)}}  Answer:  y =  frac{1 + e^{2(x + c)}}{1 - e^{2(x + c)}} + x + 2  .

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