# Thread: Integration(differential equations and first order homogenous differentail equations)

1. ## Integration(differential equations and first order homogenous differentail equations)

Here's my unsolved questions, alas, there was many questions that I can't managed to solve it. Please show me the steps and don't just give me the answer. Thanks in advance.

Expressing y in terms of x.
(1) (y/ sec^2 x)dy/dx = 2tan x + 1
(2) dy/dx = 6/(x^2 -9)
(3) (sin^2 x)dy/dx = (cos x)/y
(4) (1 - x^2)dy/dx = y(1+x^2)
(5) (y)dy/dx = √(1-y^2)
(6) (1/x)dy/dx = y^2 sec^2 x

(7) By substitution y = vx , show that x(x+y)dy/dx = x^2 + y^2 is y + x ln ((x-y)^2 /x) = 0 when y = 0 and x = 1.
(8) By substitution u = 4x + y, find the particular solution of the differential equation dy/dx = 4x + y when y = 1 and x = 0.
(9) By substitution z = 2x - 3y, show that (2x - 3y + 3)dy/dx = 2x - 3y + 1 is 2ln ( 3 - 2x + 3y ) = 1 - x + y when y = 0 and x = 1.

Please take your time, I'm having hard time solving these problems, again, thank you very much.

2. ## Re: Integration(differential equations and first order homogenous differentail equati

Alexander

I have noticed that most of them are simple D.E with separable variables.
first step separate the variables left the y right the x
then integrate it is not difficult
example get the second one dy/dx =6/(x^2-9)
after you separate the variables you must get dy=(6/x^2-9)dx
and after integration you will get y= ln|x-3|-ln|x+3| +C ,where c is the constant of integration...

good luck with the remaining.

MINOAS

Minoas

3. ## Re: Integration(differential equations and first order homogenous differentail equati

Originally Posted by alexander9408
Here's my unsolved questions, alas, there was many questions that I can't managed to solve it. Please show me the steps and don't just give me the answer. Thanks in advance.

Expressing y in terms of x.
(1) (y/ sec^2 x)dy/dx = 2tan x + 1
(2) dy/dx = 6/(x^2 -9)
(3) (sin^2 x)dy/dx = (cos x)/y
(4) (1 - x^2)dy/dx = y(1+x^2)
(5) (y)dy/dx = √(1-y^2)
(6) (1/x)dy/dx = y^2 sec^2 x

(7) By substitution y = vx , show that x(x+y)dy/dx = x^2 + y^2 is y + x ln ((x-y)^2 /x) = 0 when y = 0 and x = 1.
(8) By substitution u = 4x + y, find the particular solution of the differential equation dy/dx = 4x + y when y = 1 and x = 0.
(9) By substitution z = 2x - 3y, show that (2x - 3y + 3)dy/dx = 2x - 3y + 1 is 2ln ( 3 - 2x + 3y ) = 1 - x + y when y = 0 and x = 1.

Please take your time, I'm having hard time solving these problems, again, thank you very much.
1.
\displaystyle \begin{align*} \frac{y}{\sec^2{(x)}}\,\frac{dy}{dx} &= 2\tan{(x)} + 1 \\ y\,\frac{dy}{dx} &= \sec^2{(x)}\left[ 2\tan{(x)} + 1 \right] \\ \int{y\,\frac{dy}{dx}\,dx} &= \int{\sec^2{(x)}\left[ 2\tan{(x)} + 1 \right] dx} \\ \int{y\,dy} &= \frac{1}{2}\int{u\,du} \textrm{ after making the substitution } u = 2\tan{(x)} + 1 \implies du = 2\sec^2{(x)}\,dx \end{align*}

Go from here.

2. Solve by direct integration. You may need partial fractions.

3.
\displaystyle \begin{align*} \sin^2{(x)}\,\frac{dy}{dx} &= \frac{\cos{(x)}}{y} \\ y\,\frac{dy}{dx} &= \frac{\cos{(x)}}{\sin^2{(x)}} \\ \int{y\,\frac{dy}{dx}\,dx} &= \int{\frac{\cos{(x)}}{\sin^2{(x)}}\,dx} \\ \int{y\,dy} &= \int{u^{-2}\,du} \textrm{ after making the substitution } u = \sin{(x)} \implies du = \cos{(x)}\,dx \end{align*}

Go from here.

4.
\displaystyle \begin{align*} \left( 1 - x^2 \right) \frac{dy}{dx} &= y \left( 1 + x^2 \right) \\ \frac{1}{y}\,\frac{dy}{dx} &= \frac{x^2 + 1}{-x^2 + 1} \\ \frac{1}{y} \,\frac{dy}{dx} &= - \left( \frac{x^2 + 1}{x^2 - 1} \right) \\ \frac{1}{y}\,\frac{dy}{dx} &= -\left( 1 - \frac{2}{x^2 - 1} \right) \\ \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} &= \int{-\left( 1 - \frac{2}{x^2 - 1} \right) dx } \\ \int{\frac{1}{y}\,dy} &= \int{ -1 + \frac{2}{x^2 - 1}\,dx} \end{align*}

Go from here. You might need partial fractions to evaluate the second integral.

5.
\displaystyle \begin{align*} y\,\frac{dy}{dx} &= \sqrt{ 1 - y^2} \\ \frac{y}{\sqrt{1 - y^2}}\,\frac{dy}{dx} &= 1 \\ \int{\frac{y}{\sqrt{1-y^2}}\,\frac{dy}{dx}\,dx} &= \int{1\,dx} \\ \int{\frac{y}{\sqrt{1-y^2}}\,dy} &= \int{1\,dx} \\ -\frac{1}{2} \int{\frac{-2y}{\sqrt{1-y^2}}\,dy} &= \int{1\,dx} \\ -\frac{1}{2}\int{u^{-\frac{1}{2}}\,du} &= \int{1\,dx} \textrm{ after making the substitution } u = 1 - y^2 \implies du = -2y\,dy \end{align*}

Go from here.

6.
\displaystyle \begin{align*} \frac{1}{x}\,\frac{dy}{dx} &= y^2\sec^2{(x)} \\ y^{-2}\,\frac{dy}{dx} &= x\sec^2{(x)} \\ \int{y^{-2}\,\frac{dy}{dx}\,dx} &= \int{x\sec^2{(x)}\,dx} \\ \int{y^{-2}\,dy} &= x\tan{(x)} - \int{\tan{(x)}\,dx} \textrm{ after using integration by parts} \end{align*}

Go from here. You might need to use $\tan{(x)} = \frac{\sin{(x)}}{\cos{(x)}}$ and a substitution.