Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Certainly! Let's solve the given differential equations step-by-step.
### Part (a)
To solve the differential equation
[tex]\[ \left(x^2 + xy\right) \frac{dy}{dx} = xy + y^2, \][/tex]
we can start by rewriting it in the form:
[tex]\[ (x^2 + xy) \frac{dy}{dx} = y(x + y). \][/tex]
To solve it, we divide both sides by [tex]\(y(x+y)\)[/tex]:
[tex]\[ \frac{(x^2 + xy)}{y(x+y)} \frac{dy}{dx} = 1. \][/tex]
Simplify the left-hand side:
[tex]\[ \frac{x(x + y)}{y(x + y)} \frac{dy}{dx} = 1. \][/tex]
This simplifies to:
[tex]\[ \frac{x}{y} \frac{dy}{dx} = 1. \][/tex]
Now rearrange it to separate the variables:
[tex]\[ \frac{dy}{y} = \frac{dx}{x}. \][/tex]
Integrate both sides:
[tex]\[ \int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx, \][/tex]
which results in:
[tex]\[ \ln|y| = \ln|x| + C, \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
Exponentiate both sides:
[tex]\[ y = Cx, \][/tex]
where [tex]\(C\)[/tex] can be represented as [tex]\(C_1 = e^C\)[/tex]. Thus, the general solution to the differential equation is:
[tex]\[ y = C_1 x. \][/tex]
This is the solution to part (a).
### Part (b)
To solve the differential equation
[tex]\[ \frac{d^2 y}{dx^2} - 9y = 7e^{5x}, \][/tex]
with initial conditions [tex]\(y(0) = 2\)[/tex] and [tex]\(\frac{dy}{dx}\bigg|_{x=0} = 3\)[/tex], we need to find the general solution to the non-homogeneous differential equation.
First, find the complementary (homogeneous) solution. Consider the homogeneous equation:
[tex]\[ \frac{d^2 y}{dx^2} - 9y = 0. \][/tex]
Assume a solution of the form [tex]\(y = e^{rx}\)[/tex] and substitute it into the differential equation:
[tex]\[ r^2 e^{rx} - 9 e^{rx} = 0. \][/tex]
Factor out [tex]\(e^{rx}\)[/tex]:
[tex]\[ (r^2 - 9)e^{rx} = 0. \][/tex]
This gives us the characteristic equation:
[tex]\[ r^2 - 9 = 0, \][/tex]
[tex]\[ r^2 = 9, \][/tex]
[tex]\[ r = \pm 3. \][/tex]
So, the complementary (homogeneous) solution is:
[tex]\[ y_h = C_1 e^{3x} + C_2 e^{-3x}. \][/tex]
Next, find a particular solution to the non-homogeneous differential equation. We use the method of undetermined coefficients. Assume a particular solution of the form:
[tex]\[ y_p = A e^{5x}. \][/tex]
Substitute [tex]\(y_p\)[/tex] into the non-homogeneous differential equation:
[tex]\[ \frac{d^2}{dx^2} (Ae^{5x}) - 9(Ae^{5x}) = 7e^{5x}, \][/tex]
[tex]\[ 25Ae^{5x} - 9Ae^{5x} = 7e^{5x}, \][/tex]
[tex]\[ (25A - 9A)e^{5x} = 7e^{5x}, \][/tex]
[tex]\[ 16A e^{5x} = 7e^{5x}, \][/tex]
[tex]\[ A = \frac{7}{16}. \][/tex]
Thus, the particular solution is:
[tex]\[ y_p = \frac{7}{16} e^{5x}. \][/tex]
Now, combine the complementary solution and the particular solution:
[tex]\[ y = y_h + y_p, \][/tex]
[tex]\[ y = C_1 e^{3x} + C_2 e^{-3x} + \frac{7}{16} e^{5x}. \][/tex]
Use the initial conditions to find [tex]\(C_1\)[/tex] and [tex]\(C_2\)[/tex]. For [tex]\(y(0) = 2\)[/tex]:
[tex]\[ 2 = C_1 e^{0} + C_2 e^{0} + \frac{7}{16} e^{0}, \][/tex]
[tex]\[ 2 = C_1 + C_2 + \frac{7}{16}, \][/tex]
[tex]\[ C_1 + C_2 = 2 - \frac{7}{16}, \][/tex]
[tex]\[ C_1 + C_2 = \frac{32}{16} - \frac{7}{16}, \][/tex]
[tex]\[ C_1 + C_2 = \frac{25}{16}. \][/tex]
For [tex]\(\frac{dy}{dx}\bigg|_{x=0} = 3\)[/tex]:
[tex]\[ \frac{dy}{dx} = 3C_1 e^{3x} - 3C_2 e^{-3x} + \frac{35}{16} e^{5x}, \][/tex]
[tex]\[ \frac{dy}{dx}\bigg|_{x=0} = 3C_1 - 3C_2 + \frac{35}{16}, \][/tex]
[tex]\[ 3 = 3C_1 - 3C_2 + \frac{35}{16}, \][/tex]
[tex]\[ 3C_1 - 3C_2 = 3 - \frac{35}{16}, \][/tex]
[tex]\[ 3C_1 - 3C_2 = \frac{48}{16} - \frac{35}{16}, \][/tex]
[tex]\[ 3C_1 - 3C_2 = \frac{13}{16}. \][/tex]
Divide both equations by 3:
[tex]\[ C_1 - C_2 = \frac{13}{48}. \][/tex]
Now solve the system of linear equations:
[tex]\[ C_1 + C_2 = \frac{25}{16}, \][/tex]
[tex]\[ C_1 - C_2 = \frac{13}{48}. \][/tex]
Add the two equations:
[tex]\[ 2C_1 = \frac{25}{16} + \frac{13}{48}, \][/tex]
[tex]\[ 2C_1 = \frac{75}{48} + \frac{13}{48}, \][/tex]
[tex]\[ 2C_1 = \frac{88}{48}, \][/tex]
[tex]\[ C_1 = \frac{44}{48}, \][/tex]
[tex]\[ C_1 = \frac{11}{12}. \][/tex]
Subtract the second equation from the first:
[tex]\[ 2C_2 = \frac{25}{16} - \frac{13}{48}, \][/tex]
[tex]\[ 2C_2 = \frac{75}{48} - \frac{13}{48}, \][/tex]
[tex]\[ 2C_2 = \frac{62}{48}, \][/tex]
[tex]\[ C_2 = \frac{31}{48}. \][/tex]
So, the particular solution with initial conditions is:
[tex]\[ y = \frac{7}{16} e^{5x} + \frac{11}{12} e^{3x} + \frac{31}{48} e^{-3x}. \][/tex]
Therefore, the solution to part (b) is:
[tex]\[ y = \frac{7}{16}e^{5x} + \frac{11}{12}e^{3x} + \frac{31}{48}e^{-3x}. \][/tex]
### Final Answer
Our complete solutions are:
- Part (a): [tex]\( y(x) = C_1 x \)[/tex],
- Part (b): [tex]\( y(x) = \frac{7}{16}e^{5x} + \frac{11}{12}e^{3x} + \frac{31}{48}e^{-3x} \)[/tex].
### Part (a)
To solve the differential equation
[tex]\[ \left(x^2 + xy\right) \frac{dy}{dx} = xy + y^2, \][/tex]
we can start by rewriting it in the form:
[tex]\[ (x^2 + xy) \frac{dy}{dx} = y(x + y). \][/tex]
To solve it, we divide both sides by [tex]\(y(x+y)\)[/tex]:
[tex]\[ \frac{(x^2 + xy)}{y(x+y)} \frac{dy}{dx} = 1. \][/tex]
Simplify the left-hand side:
[tex]\[ \frac{x(x + y)}{y(x + y)} \frac{dy}{dx} = 1. \][/tex]
This simplifies to:
[tex]\[ \frac{x}{y} \frac{dy}{dx} = 1. \][/tex]
Now rearrange it to separate the variables:
[tex]\[ \frac{dy}{y} = \frac{dx}{x}. \][/tex]
Integrate both sides:
[tex]\[ \int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx, \][/tex]
which results in:
[tex]\[ \ln|y| = \ln|x| + C, \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
Exponentiate both sides:
[tex]\[ y = Cx, \][/tex]
where [tex]\(C\)[/tex] can be represented as [tex]\(C_1 = e^C\)[/tex]. Thus, the general solution to the differential equation is:
[tex]\[ y = C_1 x. \][/tex]
This is the solution to part (a).
### Part (b)
To solve the differential equation
[tex]\[ \frac{d^2 y}{dx^2} - 9y = 7e^{5x}, \][/tex]
with initial conditions [tex]\(y(0) = 2\)[/tex] and [tex]\(\frac{dy}{dx}\bigg|_{x=0} = 3\)[/tex], we need to find the general solution to the non-homogeneous differential equation.
First, find the complementary (homogeneous) solution. Consider the homogeneous equation:
[tex]\[ \frac{d^2 y}{dx^2} - 9y = 0. \][/tex]
Assume a solution of the form [tex]\(y = e^{rx}\)[/tex] and substitute it into the differential equation:
[tex]\[ r^2 e^{rx} - 9 e^{rx} = 0. \][/tex]
Factor out [tex]\(e^{rx}\)[/tex]:
[tex]\[ (r^2 - 9)e^{rx} = 0. \][/tex]
This gives us the characteristic equation:
[tex]\[ r^2 - 9 = 0, \][/tex]
[tex]\[ r^2 = 9, \][/tex]
[tex]\[ r = \pm 3. \][/tex]
So, the complementary (homogeneous) solution is:
[tex]\[ y_h = C_1 e^{3x} + C_2 e^{-3x}. \][/tex]
Next, find a particular solution to the non-homogeneous differential equation. We use the method of undetermined coefficients. Assume a particular solution of the form:
[tex]\[ y_p = A e^{5x}. \][/tex]
Substitute [tex]\(y_p\)[/tex] into the non-homogeneous differential equation:
[tex]\[ \frac{d^2}{dx^2} (Ae^{5x}) - 9(Ae^{5x}) = 7e^{5x}, \][/tex]
[tex]\[ 25Ae^{5x} - 9Ae^{5x} = 7e^{5x}, \][/tex]
[tex]\[ (25A - 9A)e^{5x} = 7e^{5x}, \][/tex]
[tex]\[ 16A e^{5x} = 7e^{5x}, \][/tex]
[tex]\[ A = \frac{7}{16}. \][/tex]
Thus, the particular solution is:
[tex]\[ y_p = \frac{7}{16} e^{5x}. \][/tex]
Now, combine the complementary solution and the particular solution:
[tex]\[ y = y_h + y_p, \][/tex]
[tex]\[ y = C_1 e^{3x} + C_2 e^{-3x} + \frac{7}{16} e^{5x}. \][/tex]
Use the initial conditions to find [tex]\(C_1\)[/tex] and [tex]\(C_2\)[/tex]. For [tex]\(y(0) = 2\)[/tex]:
[tex]\[ 2 = C_1 e^{0} + C_2 e^{0} + \frac{7}{16} e^{0}, \][/tex]
[tex]\[ 2 = C_1 + C_2 + \frac{7}{16}, \][/tex]
[tex]\[ C_1 + C_2 = 2 - \frac{7}{16}, \][/tex]
[tex]\[ C_1 + C_2 = \frac{32}{16} - \frac{7}{16}, \][/tex]
[tex]\[ C_1 + C_2 = \frac{25}{16}. \][/tex]
For [tex]\(\frac{dy}{dx}\bigg|_{x=0} = 3\)[/tex]:
[tex]\[ \frac{dy}{dx} = 3C_1 e^{3x} - 3C_2 e^{-3x} + \frac{35}{16} e^{5x}, \][/tex]
[tex]\[ \frac{dy}{dx}\bigg|_{x=0} = 3C_1 - 3C_2 + \frac{35}{16}, \][/tex]
[tex]\[ 3 = 3C_1 - 3C_2 + \frac{35}{16}, \][/tex]
[tex]\[ 3C_1 - 3C_2 = 3 - \frac{35}{16}, \][/tex]
[tex]\[ 3C_1 - 3C_2 = \frac{48}{16} - \frac{35}{16}, \][/tex]
[tex]\[ 3C_1 - 3C_2 = \frac{13}{16}. \][/tex]
Divide both equations by 3:
[tex]\[ C_1 - C_2 = \frac{13}{48}. \][/tex]
Now solve the system of linear equations:
[tex]\[ C_1 + C_2 = \frac{25}{16}, \][/tex]
[tex]\[ C_1 - C_2 = \frac{13}{48}. \][/tex]
Add the two equations:
[tex]\[ 2C_1 = \frac{25}{16} + \frac{13}{48}, \][/tex]
[tex]\[ 2C_1 = \frac{75}{48} + \frac{13}{48}, \][/tex]
[tex]\[ 2C_1 = \frac{88}{48}, \][/tex]
[tex]\[ C_1 = \frac{44}{48}, \][/tex]
[tex]\[ C_1 = \frac{11}{12}. \][/tex]
Subtract the second equation from the first:
[tex]\[ 2C_2 = \frac{25}{16} - \frac{13}{48}, \][/tex]
[tex]\[ 2C_2 = \frac{75}{48} - \frac{13}{48}, \][/tex]
[tex]\[ 2C_2 = \frac{62}{48}, \][/tex]
[tex]\[ C_2 = \frac{31}{48}. \][/tex]
So, the particular solution with initial conditions is:
[tex]\[ y = \frac{7}{16} e^{5x} + \frac{11}{12} e^{3x} + \frac{31}{48} e^{-3x}. \][/tex]
Therefore, the solution to part (b) is:
[tex]\[ y = \frac{7}{16}e^{5x} + \frac{11}{12}e^{3x} + \frac{31}{48}e^{-3x}. \][/tex]
### Final Answer
Our complete solutions are:
- Part (a): [tex]\( y(x) = C_1 x \)[/tex],
- Part (b): [tex]\( y(x) = \frac{7}{16}e^{5x} + \frac{11}{12}e^{3x} + \frac{31}{48}e^{-3x} \)[/tex].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
