1. Express (1)/(x(2x+3) ) Is partial factions. 2. The variables x and y satisfy the differential equation. x(2x + 3) (dy)/(dx ) = y and it is given that y = 1 when x = 1. a. fin…

Question

1. Express
`(1)/(x(2x+3) )` Is partial factions.

2. The variables x and y satisfy the differential equation.
x(2x + 3)
`(dy)/(dx )` = y and it is given that y = 1 when x = 1.
a.
find the particular solution of above differential equation, express y in term of x.
b.
Calculate the value of y when x = 9. given your answer in two decimal places.

Answer 1

1. Let a and b be coefficients such that

`\frac1{x(2x+3)} = \frac ax + \frac b{2x+3}`

Combining the fractions on the right gives

`\frac1{x(2x+3)} = (a(2x+3) + bx)/(x(2x+3))`

`\implies 1 = (2a+b)x + 3a`

`\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\frac13, b = -\frac23`

so that

`\frac1{x(2x+3)} = \boxed{\frac13 \left(\frac1x - \frac2{2x+3}\right)}`

2. a. The given ODE is separable as

`x(2x+3) \frac{dy}dx} = y \implies \frac{dy}y = (dx)/(x(2x+3))`

Using the result of part (1), integrating both sides gives

`\ln|y| = \frac13 \left(\ln|x| - \ln|2x+3|\right) + C`

Given that y = 1 when x = 1, we find

`\ln|1| = \frac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \frac13\ln(5)`

so the particular solution to the ODE is

`\ln|y| = \frac13 \left(\ln|x| - \ln|2x+3|\right) + \frac13\ln(5)`

We can solve this explicitly for y :

`\ln|y| = \frac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)`

`\ln|y| = \frac13 \ln\left|(5x)/(2x+3)\right|`

`\ln|y| = \ln\left|\sqrt[3]{(5x)/(2x+3)}\right|`

`\boxed{y = \sqrt[3]{(5x)/(2x+3)}}`

2. b. When x = 9, we get

`y = \sqrt[3]{(45)/(21)} = \sqrt[3]{\frac{15}7} \approx \boxed{1.29}`

Answer 2

`\textsf{1.} \quad (1)/(x(2x+3)) \equiv (1)/(3x)-(2)/(3(2x+3))`

`\textsf{2a.} \quad y= \sqrt[3]{(5x)/(2x+3)}`

`\textsf{2b.} \quad 1.29`

Explanation:

Question 1

A fraction with more than one linear factor in the denominator can be decomposed into partial fractions. This means writing it as a sum of two or more simpler fractions, making it easier to integrate, manipulate, or solve equations involving the original fraction.

Given fraction:

`(1)/(x(2x+3))`

Write out the expression as an identity:

`(1)/(x(2x+3)) \equiv (A)/(x)+(B)/(2x+3)`

Add the partial fractions:

`(1)/(x(2x+3)) \equiv (A(2x+3)+Bx)/(x(2x+3))`

Cancel the denominators from both sides of the identity, so the numerators are equal:

`1 \equiv A(2x+3)+Bx`

Rearrange:

`1 \equiv 2Ax+3A+Bx`

`1 \equiv (2A+B)x+3A`

Solve the unknown parameters A and B by plugging in the real roots of the denominator of the original fraction: x = 0 and x = -3/2.

`\begin{aligned}x=0 \implies 1 &=(2A+B)(0)+3A\\1&=3A\\A&=(1)/(3)\end{aligned}`

`\begin{aligned}x=-(3)/(2) \implies 1 &=(2A+B)\left(-(3)/(2)\right)+3A\\\\A=(1)/(3) \implies 1&=\left(2\left((1)/(3)\right)+B\right)\left(-(3)/(2)\right)+3\left((1)/(3)\right)\\\\1&=-1-(3)/(2)B+1\\\\1&=-(3)/(2)B\\\\B&=-(2)/(3)\end{aligned}`

Finally, replace A and B in the original identity to express the given fraction as partial fractions:

`\boxed{(1)/(x(2x+3)) \equiv (1)/(3x)-(2)/(3(2x+3))}`

`\hrulefill`

Question 2a

Given differential equation:

`x(2x + 3) \frac{\text{d}y}{\text{d}x}= y`

To solve the differential equation, first move all the terms containing y to the left-hand side, and all the terms containing x to the right-hand side, and split up the dy/dx:

`(1)/(y)\:\text{d}y= (1)/(x(2x + 3))\:\text{d}x`

Now integrate both sides:

`\displaystyle \int (1)/(y)\:\text{d}y= \int (1)/(x(2x + 3))\:\text{d}x`

Substitute the found partial fractions from the previous question:

`\displaystyle \int (1)/(y)\:\text{d}y= \int \left((1)/(3x)-(2)/(3(2x+3))\right)\:\text{d}x`

Simplify by separating the integrals and taking the constants outside:

`\displaystyle \int (1)/(y)\:\text{d}y= (1)/(3)\int (1)/(x)\:\text{d}x- (2)/(3)\int(1)/(2x+3)\right)\:\text{d}x`

Integrate:

`\ln|y|=(1)/(3)\ln|x|-(2)/(3)\cdot (1)/(2)\ln|2x+3|+C`

`\ln|y|=(1)/(3)\ln|x|- (1)/(3)\ln|2x+3|+C`

To find the value of C, substitute the given parameters of y = 1 when x = 1:

`\ln|1|=(1)/(3)\ln|1|- (1)/(3)\ln|2(1)+3|+C`

`0=(1)/(3)(0)- (1)/(3)\ln5+C`

`C=(1)/(3)\ln5`

Therefore:

`\ln|y|=(1)/(3)\ln|x|- (1)/(3)\ln|2x+3|+(1)/(3)\ln5`

Factor out 1/3 from the right side of the equation:

`\ln|y|=(1)/(3)\left(\ln|x|-\ln|2x+3|+\ln5\right)`

Use log rules to simplify the right side of the equation:

`\ln|y|=(1)/(3)\left(\ln \left|(x)/(2x+3) \right|+\ln5\right)`

`\ln|y|=(1)/(3)\ln \left|(5x)/(2x+3) \right|`

`\ln|y|=\ln \sqrt[3]\left`

Therefore:

`y= \sqrt[3]{(5x)/(2x+3)}`

`\hrulefill`

Question 2b

To calculate the value of y when x = 9, substitute x = 9 into the equation for y found in question 2a:

`\begin{aligned}x=9 \implies y&= \sqrt[3]{(5(9))/(2(9)+3)}\\\\&= \sqrt[3]{(45)/(18+3)}\\\\&= \sqrt[3]{(45)/(21)}\\\\&= \sqrt[3]{(15 \cdot 3)/(7 \cdot 3)}\\\\&= \sqrt[3]{(15)/(7)}\\\\&= 1.28923198...\\\\&=1.29\; \sf (2\;d.p.)\end{aligned}`

Therefore, the value of y when x = 9 is 1.29 to two decimal places.