Solved on Mar 14, 2024

Find the solution to the ODE $y' = -1.7xy$ with $y(0) = 9.6$ , and evaluate the solution at $x = 1.3$ .

STEP 1

Assumptions
1. The differential equation is given by $y' = -1.7xy$ .
2. The initial condition is given by $y(0) = 9.6$ .
3. We are looking for the solution $y(x)$ that satisfies both the differential equation and the initial condition.
4. We will use separation of variables to solve the differential equation.

STEP 2

Write the differential equation in a form that allows for separation of variables.
$y' = -1.7xy$

STEP 3

Separate the variables by dividing both sides by $y$ and then by $x$ (assuming $y \neq 0$ and $x \neq 0$ ).
$\frac{1}{y} dy = -1.7x dx$

STEP 4

Integrate both sides of the equation with respect to their respective variables.
$\int \frac{1}{y} dy = \int -1.7x dx$

STEP 5

Perform the integration on both sides.
$\ln|y| = -0.85x^2 + C$

STEP 6

Exponentiate both sides to solve for $y$ .
$e^{\ln|y|} = e^{-0.85x^2 + C}$

STEP 7

Simplify the left side and use the property that $e^{A+B} = e^A \cdot e^B$ on the right side.
$|y| = e^{-0.85x^2} \cdot e^C$

STEP 8

Since $e^C$ is just a constant, we can denote it as $C_1$ where $C_1 = e^C > 0$ .
$|y| = C_1 e^{-0.85x^2}$

STEP 9

Remove the absolute value by considering that $y$ can be either positive or negative, which means $y = \pm C_1 e^{-0.85x^2}$ .

STEP 10

Apply the initial condition $y(0) = 9.6$ to determine the value of $C_1$ .
$9.6 = \pm C_1 e^{-0.85(0)^2}$

STEP 11

Since $e^0 = 1$ , we have:
$9.6 = \pm C_1$

STEP 12

Since $C_1 > 0$ and $y(0) = 9.6$ is positive, we choose the positive value for $C_1$ .
$C_1 = 9.6$

STEP 13

Write the particular solution with the determined constant.
$y(x) = 9.6 e^{-0.85x^2}$

STEP 14

Evaluate the solution at $x = 1.3$ .
$y(1.3) = 9.6 e^{-0.85(1.3)^2}$

STEP 15

Calculate the exponent $-0.85(1.3)^2$ .
$-0.85(1.3)^2 = -1.4305$

STEP 16

Evaluate the exponential function.
$y(1.3) = 9.6 e^{-1.4305}$

STEP 17

Use a calculator to find the value of $e^{-1.4305}$ to at least three decimal places.
$e^{-1.4305} \approx 0.239$

STEP 18

Multiply the result by $9.6$ .
$y(1.3) = 9.6 \times 0.239$

STEP 19

Calculate the final result.
$y(1.3) \approx 9.6 \times 0.239 \approx 2.2944$
The value of the solution at $x = 1.3$ is approximately $2.294$ to three decimal places.

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Find the solution to the ODE $y' = -1.7xy$ with $y(0) = 9.6$ , and evaluate the solution at $x = 1.3$ .

STEP 1

STEP 2

Write the differential equation in a form that allows for separation of variables.
$y' = -1.7xy$

STEP 3

Separate the variables by dividing both sides by $y$ and then by $x$ (assuming $y \neq 0$ and $x \neq 0$ ).
$\frac{1}{y} dy = -1.7x dx$

STEP 4

Integrate both sides of the equation with respect to their respective variables.
$\int \frac{1}{y} dy = \int -1.7x dx$

STEP 5

Perform the integration on both sides.
$\ln|y| = -0.85x^2 + C$

STEP 6

Exponentiate both sides to solve for $y$ .
$e^{\ln|y|} = e^{-0.85x^2 + C}$

STEP 7

Simplify the left side and use the property that $e^{A+B} = e^A \cdot e^B$ on the right side.
$|y| = e^{-0.85x^2} \cdot e^C$

STEP 8

Since $e^C$ is just a constant, we can denote it as $C_1$ where $C_1 = e^C > 0$ .
$|y| = C_1 e^{-0.85x^2}$

STEP 9

Remove the absolute value by considering that $y$ can be either positive or negative, which means $y = \pm C_1 e^{-0.85x^2}$ .

STEP 10

Apply the initial condition $y(0) = 9.6$ to determine the value of $C_1$ .
$9.6 = \pm C_1 e^{-0.85(0)^2}$

STEP 11

Since $e^0 = 1$ , we have:
$9.6 = \pm C_1$

STEP 12

Since $C_1 > 0$ and $y(0) = 9.6$ is positive, we choose the positive value for $C_1$ .
$C_1 = 9.6$

STEP 13

Write the particular solution with the determined constant.
$y(x) = 9.6 e^{-0.85x^2}$

STEP 14

Evaluate the solution at $x = 1.3$ .
$y(1.3) = 9.6 e^{-0.85(1.3)^2}$

STEP 15

Calculate the exponent $-0.85(1.3)^2$ .
$-0.85(1.3)^2 = -1.4305$

STEP 16

Evaluate the exponential function.
$y(1.3) = 9.6 e^{-1.4305}$

STEP 17

Use a calculator to find the value of $e^{-1.4305}$ to at least three decimal places.
$e^{-1.4305} \approx 0.239$

STEP 18

Multiply the result by $9.6$ .
$y(1.3) = 9.6 \times 0.239$

STEP 19

Calculate the final result.
$y(1.3) \approx 9.6 \times 0.239 \approx 2.2944$
The value of the solution at $x = 1.3$ is approximately $2.294$ to three decimal places.

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Find the solution to the ODE y′=−1.7xyy' = -1.7xyy′=−1.7xy with y(0)=9.6y(0) = 9.6y(0)=9.6, and evaluate the solution at x=1.3x = 1.3x=1.3.

STEP 1

STEP 2

Write the differential equation in a form that allows for separation of variables.y′=−1.7xy y' = -1.7xy y′=−1.7xy

STEP 3

Separate the variables by dividing both sides by y y y and then by x x x (assuming y≠0 y \neq 0 y=0 and x≠0 x \neq 0 x=0).1ydy=−1.7xdx \frac{1}{y} dy = -1.7x dx y1​dy=−1.7xdx

STEP 4

Integrate both sides of the equation with respect to their respective variables.∫1ydy=∫−1.7xdx \int \frac{1}{y} dy = \int -1.7x dx ∫y1​dy=∫−1.7xdx

STEP 5

Perform the integration on both sides.ln⁡∣y∣=−0.85x2+C \ln|y| = -0.85x^2 + C ln∣y∣=−0.85x2+C

STEP 6

Exponentiate both sides to solve for y y y.eln⁡∣y∣=e−0.85x2+C e^{\ln|y|} = e^{-0.85x^2 + C} eln∣y∣=e−0.85x2+C

STEP 7

Simplify the left side and use the property that eA+B=eA⋅eB e^{A+B} = e^A \cdot e^B eA+B=eA⋅eB on the right side.∣y∣=e−0.85x2⋅eC |y| = e^{-0.85x^2} \cdot e^C ∣y∣=e−0.85x2⋅eC

STEP 8

Since eC e^C eC is just a constant, we can denote it as C1 C_1 C1​ where C1=eC>0 C_1 = e^C > 0 C1​=eC>0.∣y∣=C1e−0.85x2 |y| = C_1 e^{-0.85x^2} ∣y∣=C1​e−0.85x2

STEP 9

Remove the absolute value by considering that y y y can be either positive or negative, which means y=±C1e−0.85x2 y = \pm C_1 e^{-0.85x^2} y=±C1​e−0.85x2.

STEP 10

Apply the initial condition y(0)=9.6 y(0) = 9.6 y(0)=9.6 to determine the value of C1 C_1 C1​.9.6=±C1e−0.85(0)2 9.6 = \pm C_1 e^{-0.85(0)^2} 9.6=±C1​e−0.85(0)2

STEP 11

Since e0=1 e^0 = 1 e0=1, we have:9.6=±C1 9.6 = \pm C_1 9.6=±C1​

STEP 12

Since C1>0 C_1 > 0 C1​>0 and y(0)=9.6 y(0) = 9.6 y(0)=9.6 is positive, we choose the positive value for C1 C_1 C1​.C1=9.6 C_1 = 9.6 C1​=9.6

STEP 13

Write the particular solution with the determined constant.y(x)=9.6e−0.85x2 y(x) = 9.6 e^{-0.85x^2} y(x)=9.6e−0.85x2

STEP 14

Evaluate the solution at x=1.3 x = 1.3 x=1.3.y(1.3)=9.6e−0.85(1.3)2 y(1.3) = 9.6 e^{-0.85(1.3)^2} y(1.3)=9.6e−0.85(1.3)2

STEP 15

Calculate the exponent −0.85(1.3)2 -0.85(1.3)^2 −0.85(1.3)2.−0.85(1.3)2=−1.4305 -0.85(1.3)^2 = -1.4305 −0.85(1.3)2=−1.4305

STEP 16

Evaluate the exponential function.y(1.3)=9.6e−1.4305 y(1.3) = 9.6 e^{-1.4305} y(1.3)=9.6e−1.4305

STEP 17

Use a calculator to find the value of e−1.4305 e^{-1.4305} e−1.4305 to at least three decimal places.e−1.4305≈0.239 e^{-1.4305} \approx 0.239 e−1.4305≈0.239

STEP 18

Multiply the result by 9.6 9.6 9.6.y(1.3)=9.6×0.239 y(1.3) = 9.6 \times 0.239 y(1.3)=9.6×0.239

STEP 19

Calculate the final result.y(1.3)≈9.6×0.239≈2.2944 y(1.3) \approx 9.6 \times 0.239 \approx 2.2944 y(1.3)≈9.6×0.239≈2.2944The value of the solution at x=1.3 x = 1.3 x=1.3 is approximately 2.294 2.294 2.294 to three decimal places.

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Find the solution to the ODE $y' = -1.7xy$ with $y(0) = 9.6$ , and evaluate the solution at $x = 1.3$ .

Write the differential equation in a form that allows for separation of variables.
$y' = -1.7xy$

Separate the variables by dividing both sides by $y$ and then by $x$ (assuming $y \neq 0$ and $x \neq 0$ ).
$\frac{1}{y} dy = -1.7x dx$

Integrate both sides of the equation with respect to their respective variables.
$\int \frac{1}{y} dy = \int -1.7x dx$

Perform the integration on both sides.
$\ln|y| = -0.85x^2 + C$

Exponentiate both sides to solve for $y$ .
$e^{\ln|y|} = e^{-0.85x^2 + C}$

Simplify the left side and use the property that $e^{A+B} = e^A \cdot e^B$ on the right side.
$|y| = e^{-0.85x^2} \cdot e^C$

Since $e^C$ is just a constant, we can denote it as $C_1$ where $C_1 = e^C > 0$ .
$|y| = C_1 e^{-0.85x^2}$

Remove the absolute value by considering that $y$ can be either positive or negative, which means $y = \pm C_1 e^{-0.85x^2}$ .

Apply the initial condition $y(0) = 9.6$ to determine the value of $C_1$ .
$9.6 = \pm C_1 e^{-0.85(0)^2}$

Since $e^0 = 1$ , we have:
$9.6 = \pm C_1$

Since $C_1 > 0$ and $y(0) = 9.6$ is positive, we choose the positive value for $C_1$ .
$C_1 = 9.6$

Write the particular solution with the determined constant.
$y(x) = 9.6 e^{-0.85x^2}$

Evaluate the solution at $x = 1.3$ .
$y(1.3) = 9.6 e^{-0.85(1.3)^2}$

Calculate the exponent $-0.85(1.3)^2$ .
$-0.85(1.3)^2 = -1.4305$

Evaluate the exponential function.
$y(1.3) = 9.6 e^{-1.4305}$

Use a calculator to find the value of $e^{-1.4305}$ to at least three decimal places.
$e^{-1.4305} \approx 0.239$

Multiply the result by $9.6$ .
$y(1.3) = 9.6 \times 0.239$

Calculate the final result.
$y(1.3) \approx 9.6 \times 0.239 \approx 2.2944$
The value of the solution at $x = 1.3$ is approximately $2.294$ to three decimal places.