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.

Answer 1

$$\begin{array}{l} \frac{dy}{dx} - ay = 0 \\ \frac{dy}{y} = adx \\ \ln y = ax + \ln C \\ y = C(x)e^{ax} \\ \frac{dy}{dx} = \frac{dC}{dx}e^{ax} + Cae^{ax} \\ \end{array}$$

a)

$$\begin{array}{l} \frac{dC}{dx}e^{ax} + Cae^{ax} - Cae^{ax} = ke^{\lambda(x)} \\ \frac{dC}{dx} = \frac{ke^{\lambda(x)}}{e^{ax}} \\ C(x) = \int \frac{ke^{\lambda(x)}}{e^{ax}} dx \\ \end{array}$$

Answer:

$$y = \left(\int \frac{ke^{\lambda(x)}}{e^{ax}} dx\right) * e^{ax}$$

B)

$$\begin{array}{l} \frac{dC}{dx}e^{ax} + Cae^{ax} - Cae^{ax} = ke^{ax} \\ \frac{dC}{dx} = \frac{ke^{ax}}{e^{ax}} \\ C(x) = k\int x\,dx = kx + c_1 \\ \end{array}$$

Answer:

$$y = (kx + c_1) * e^{ax}$$

c)

$$\begin{array}{l} \frac{dC}{dx}e^{ax} + Cae^{ax} - Cae^{ax} = kx^n e^{ax} \\ \frac{dC}{dx} = kx^n \\ C(x) = k\int x^n\,dx = k \frac{x^{n+1}}{n+1} + c_1 \\ \end{array}$$

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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