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2 years ago

Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

Mathematics

1 answer:

goblinko [34]2 years ago

4 0

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

$\sqrt{4\cdot 6+25}=2\cdot 1+C$

$C=5$

Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

$y(x)=x^2+5x$

Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

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2 years ago

2 u + 12 < 23 0,1 d + 8 > 0 3 - 4 r > 7 13 < 2 z + 3 6 ≥ 9 - 0, 15 i

BartSMP [9]

Answer:

Part 1) $u< 5.5$

Part 2) $d > -80$

Part 3) $r < -1$

Part 4) $z>5$

Part 5) $i\geq 20$

Step-by-step explanation:

The question is

Solve each inequality for the indicated variable

Part 1) we have

$2u+12$

subtract 12 both sides

$2u< 23-12\\2u$

Divide by 2 both sides

$u< 5.5$

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

$0.1d+8 >0$

subtract 8 both sides

$0.1d > -8$

Divide by 0.1 both sides

$d > -80$

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

$3-4r> 7$

Subtract 3 both sides

$-4r>7-3\\-4r>4$

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$r < -1$

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

$13< 2z+3$

Subtract 3 both sides

$13-3$

Divide by 2 both sides

$5$

Rewrite

$z>5$

The solution is the interval (5,∞)

In a number line the solution is the shaded area at right of z=5 (open circle)

The number 5 is not included in the solution

Part 5) we have

$6\geq 9-0.15i$

Subtract 9 both sides

$6-9\geq -0.15i\\-3\geq -0.15i$

Divide by -0.15 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$20\leq i$

Rewrite

$i\geq 20$

The solution is the interval [20,∞)

In a number line the solution is the shaded area at right of i=20 (closed circle)

The number 20 is included in the solution

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3 years ago

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yan [13]

The area of D is given by:

$\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx \\ \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}$

The average value of f over D is given by:

$\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49$

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Solution :

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When the snail crawls 1/3 of a mile per day, it will take him, $1 \times \frac{3}{1} \times 3$

= 9 days to get to the next pond.

When the snail crawls 1/4 of a mile per day, it will take him, $1 \times \frac{4}{1} \times 3$

= 12 days to get to the next pond.

When the snail crawls 3/4 of a mile per day, it will take him, $1 \times \frac{4}{3} \times 3$

= 4 days to get to the next pond.

When the snail crawls 2/3 of a mile per day, it will take him, $1 \times \frac{3}{2} \times 3$

= 4.5 days to get to the next pond.

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3 years ago