QuestionFind $k$ such that $y=\frac{1}{k x^{3}}$ solves $\frac{d y}{d x}=13 x^{2} y^{2}$ . Round to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The function $y=\frac{1}{k x^{3}}$ is a solution to the differential equation $\frac{d y}{d x}=13 x^{} y^{}$ . . We need to find the value of $k$ that makes this true.

STEP 2

First, we need to find the derivative of $y=\frac{1}{k x^{}}$ .
$\frac{d y}{d x} = -\frac{}{k x^{4}}$

STEP 3

Now, we can substitute $y=\frac{1}{k x^{3}}$ and $\frac{d y}{d x} = -\frac{3}{k x^{}}$ into the differential equation $\frac{d y}{d x}=13 x^{2} y^{2}$ .
$-\frac{3}{k x^{}} =13 x^{2} \left(\frac{1}{k x^{3}}\right)^{2}$

STEP 4

implify the equation.
$-\frac{3}{k x^{4}} =13 \left(\frac{1}{k^{2} x^{6}}\right)$

STEP 5

Multiply both sides by $-k^{2} x^{4}$ to get rid of the fractions.
$3k =13 x^{2}$

STEP 6

Since the equation should hold for all $x$ , we can choose a convenient value for $x$ . Let's choose $x=1$ .
$3k =13$

STEP 7

olve for $k$ .
$k = \frac{13}{3}$

STEP 8

Round the value of $k$ to the nearest tenth.
$k \approx4.3$ The value of $k$ for which $y=\frac{1}{k x^{3}}$ is a solution to the differential equation $\frac{d y}{d x}=13 x^{2} y^{2}$ is approximately4.3.

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QuestionFind $k$ such that $y=\frac{1}{k x^{3}}$ solves $\frac{d y}{d x}=13 x^{2} y^{2}$ . Round to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The function $y=\frac{1}{k x^{3}}$ is a solution to the differential equation $\frac{d y}{d x}=13 x^{} y^{}$ . . We need to find the value of $k$ that makes this true.

STEP 2

First, we need to find the derivative of $y=\frac{1}{k x^{}}$ .
$\frac{d y}{d x} = -\frac{}{k x^{4}}$

STEP 3

Now, we can substitute $y=\frac{1}{k x^{3}}$ and $\frac{d y}{d x} = -\frac{3}{k x^{}}$ into the differential equation $\frac{d y}{d x}=13 x^{2} y^{2}$ .
$-\frac{3}{k x^{}} =13 x^{2} \left(\frac{1}{k x^{3}}\right)^{2}$

STEP 4

implify the equation.
$-\frac{3}{k x^{4}} =13 \left(\frac{1}{k^{2} x^{6}}\right)$

STEP 5

Multiply both sides by $-k^{2} x^{4}$ to get rid of the fractions.
$3k =13 x^{2}$

STEP 6

Since the equation should hold for all $x$ , we can choose a convenient value for $x$ . Let's choose $x=1$ .
$3k =13$

STEP 7

olve for $k$ .
$k = \frac{13}{3}$

STEP 8

Round the value of $k$ to the nearest tenth.
$k \approx4.3$ The value of $k$ for which $y=\frac{1}{k x^{3}}$ is a solution to the differential equation $\frac{d y}{d x}=13 x^{2} y^{2}$ is approximately4.3.

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QuestionFind kkk such that y=1kx3y=\frac{1}{k x^{3}}y=kx31​ solves dydx=13x2y2\frac{d y}{d x}=13 x^{2} y^{2}dxdy​=13x2y2. Round to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The function y=1kx3y=\frac{1}{k x^{3}}y=kx31​ is a solution to the differential equation dydx=13xy\frac{d y}{d x}=13 x^{} y^{}dxdy​=13xy. . We need to find the value of kkk that makes this true.

STEP 2

First, we need to find the derivative of y=1kxy=\frac{1}{k x^{}}y=kx1​.dydx=−kx4\frac{d y}{d x} = -\frac{}{k x^{4}}dxdy​=−kx4​

STEP 3

STEP 4

implify the equation.−3kx4=13(1k2x6)-\frac{3}{k x^{4}} =13 \left(\frac{1}{k^{2} x^{6}}\right)−kx43​=13(k2x61​)

STEP 5

Multiply both sides by −k2x4-k^{2} x^{4}−k2x4 to get rid of the fractions.3k=13x23k =13 x^{2}3k=13x2

STEP 6

Since the equation should hold for all xxx, we can choose a convenient value for xxx. Let's choose x=1x=1x=1.3k=133k =133k=13

STEP 7

olve for kkk.k=133k = \frac{13}{3}k=313​

STEP 8

Round the value of kkk to the nearest tenth.k≈4.3k \approx4.3k≈4.3The value of kkk for which y=1kx3y=\frac{1}{k x^{3}}y=kx31​ is a solution to the differential equation dydx=13x2y2\frac{d y}{d x}=13 x^{2} y^{2}dxdy​=13x2y2 is approximately4.3.

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QuestionFind $k$ such that $y=\frac{1}{k x^{3}}$ solves $\frac{d y}{d x}=13 x^{2} y^{2}$ . Round to the nearest tenth.

Assumptions1. The function $y=\frac{1}{k x^{3}}$ is a solution to the differential equation $\frac{d y}{d x}=13 x^{} y^{}$ . . We need to find the value of $k$ that makes this true.

First, we need to find the derivative of $y=\frac{1}{k x^{}}$ .
$\frac{d y}{d x} = -\frac{}{k x^{4}}$

implify the equation.
$-\frac{3}{k x^{4}} =13 \left(\frac{1}{k^{2} x^{6}}\right)$

Multiply both sides by $-k^{2} x^{4}$ to get rid of the fractions.
$3k =13 x^{2}$

Since the equation should hold for all $x$ , we can choose a convenient value for $x$ . Let's choose $x=1$ .
$3k =13$

olve for $k$ .
$k = \frac{13}{3}$

Round the value of $k$ to the nearest tenth.
$k \approx4.3$ The value of $k$ for which $y=\frac{1}{k x^{3}}$ is a solution to the differential equation $\frac{d y}{d x}=13 x^{2} y^{2}$ is approximately4.3.