initial-value problem<\/h3>\r\nAn initial-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at a single point in time [latex]t_0[\/latex]. For second-order equations, we need two initial conditions, typically [latex]y(t_0)[\/latex] and [latex]y'(t_0)[\/latex].\r\n\r\n<\/section>Sometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[\/latex] and [latex]y(t_1)=y_1[\/latex]. These conditions are called boundary conditions<\/strong>, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem<\/strong>.\r\n\r\n\r\nboundary-value problem<\/h3>\r\nA boundary-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at two different points in time, [latex]t_0[\/latex] and [latex]t_1[\/latex].\r\n\r\n<\/section>Within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.<\/section>Solve the following initial-value problem: [latex]y''+3y'-4y=0[\/latex],\u00a0[latex]y(0)=1[\/latex],\u00a0[latex]y'(0)=-9[\/latex].[reveal-answer q=\"380210266\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380210266\"]\r\nWe already solved this differential equation previously and found the general solution to be:<\/p>\r\n
[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n
Then:<\/p>\r\n
[latex]y^\\prime=-4c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n
When [latex]x=0[\/latex], we have [latex]y(0)=c_1+c_2[\/latex] and [latex]y^\\prime(0)=-4c_1+c_2[\/latex]. Applying the initial conditions, we have:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1+c_2&=1 \\\\\r\n-4c_1+c_2&=-9\r\n\\end{aligned}[\/latex].<\/p>\r\n
Then [latex]c_1=1-c_2[\/latex]. Substituting this expression into the second equation, we see that:<\/p>\r\n
[latex]\\begin{aligned}\r\n-4(1-c_2)+c_2&=-9 \\\\\r\n-4+4c_2+c_2&=-9 \\\\\r\n5c_2&=-5 \\\\\r\nc_2&=-1\r\n\\end{aligned}[\/latex].<\/p>\r\n
So, [latex]c_1=2[\/latex] and the solution to the initial-value problem is:<\/p>\r\n
[latex]y(x)=2e^{-4x}-e^x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nSolve the following initial-value problem and graph the solution:<\/p>\r\n
[latex]y^{\\prime\\prime}+6y^\\prime+13y=0, \\ y(0)=0, \\ y^\\prime(0)=2[\/latex].<\/p>\r\n[reveal-answer q=\"835401193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"835401193\"]\r\n
We already solved this differential equation previously and found the general solution to be:<\/p>\r\n
[latex]y(x)=e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\nThen:\r\n
<\/div>\r\n[latex]y^\\prime(x)=e^{-3x}(-2c_1\\sin2x+2c_2\\cos2x)-3e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\n
When [latex]x=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y^\\prime(0)=2c_2-3c_1[\/latex]. Applying the initial conditions, we obtain:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1&=0 \\\\\r\n-3c_1+2c_2&= 2\r\n\\end{aligned}[\/latex].<\/p>\r\n
Therefore, [latex]c_1=0[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is shown in the following graph.<\/p>\r\n
[latex]y=e^{-3x}\\sin2x[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_3273\" align=\"aligncenter\" width=\"723\"]![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014029\/7-1-1.jpeg\")
Figure 1.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nThe following initial-value problem models the position of an object with mass attached to a spring. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.<\/em><\/p>\r\n[latex]y^{\\prime\\prime}+2y^\\prime+y=0, \\ y(0)=1, \\ y^\\prime(0)=0[\/latex]<\/p>\r\n
Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[\/latex] sec? How fast is the mass moving at time [latex]t=1[\/latex] sec? In what direction?<\/p>\r\n[reveal-answer q=\"394824182\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394824182\"]\r\n
We already solved this previously and found the general solution to this differential equation to be:<\/p>\r\n
[latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex].<\/p>\r\n
Then:<\/p>\r\n
[latex]y^\\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[\/latex].<\/p>\r\n
When [latex]t=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y'(0)=-c_1+c_2[\/latex]. Applying the initial conditions, we obtain:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1&=1 \\\\\r\n-c_1+c_2&= 0\r\n\\end{aligned}[\/latex].<\/p>\r\n
Thus, [latex]c_1=1[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is:<\/p>\r\n
[latex]y(t)=e^{-t}+te^{-t}[\/latex].<\/p>\r\n
This solution is represented in the following graph. At time [latex]t=2[\/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\\approx0.406[\/latex] [latex]m[\/latex] below equilibrium.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3276\" align=\"aligncenter\" width=\"702\"]![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014137\/7-1-2.jpeg\")
Figure 3.[\/caption]\r\n\r\n \r\n
To calculate the velocity at time [latex]t=1[\/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[\/latex], so:<\/p>\r\n
[latex]y^\\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[\/latex].<\/p>\r\n
Then [latex]y^\\prime(1)=e^{-1}\\approx-0.3679[\/latex]. At time [latex]t=1[\/latex], the mass is moving upward at [latex]0.3679[\/latex] m\/sec.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nEarlier we solved the differential equation [latex]y''+16y=0[\/latex] and found the general solution to be [latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:<\/p>\r\n\r\n
\r\n \t- [latex]y(0)=0[\/latex],\u00a0[latex]y\\left(\\frac{\\pi}4\\right)=0[\/latex]<\/li>\r\n \t
- [latex]y(0)=1[\/latex], [latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex]<\/li>\r\n \t
- [latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex],\u00a0[latex]y\\left(\\frac{3\\pi}8\\right)=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"437288237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437288237\"]\r\n
We have:<\/p>\r\n\r\n
\r\n[latex]y(x)=c_1\\cos{4t}+c_2\\sin{4t}[\/latex]<\/p>\r\n\r\n<\/div>\r\n
\r\n \t- Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[\/latex]. So the solution is of the form [latex]y(t)=c_2\\sin4t[\/latex]. When we apply the second boundary condition, though, we get [latex]y\\left(\\frac{\\pi}4\\right)=c_2\\sin\\left(4\\left(\\frac{\\pi}4\\right)\\right)=c_2\\sin\\pi=0[\/latex] for all values of [latex]c_2[\/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[\/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\\sin4t[\/latex] is a solution for any value of [latex]c_2[\/latex].<\/li>\r\n \t
- Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[\/latex]. Applying the second boundary condition gives[latex]y(\\frac{\\pi}{8})=c_2=0[\/latex], so [latex]c_2=0[\/latex]. In this case, we have a unique solution: [latex]y(t)=\\cos 4t[\/latex].<\/li>\r\n \t
- Applying the first boundary condition given here, we get [latex]y\\left(\\frac{\\pi}8\\right)=c_2=0[\/latex]. However, applying the second boundary condition gives [latex]y\\left(\\frac{3\\pi}8\\right)=-c_2=2[\/latex], so [latex]c_2=-2[\/latex]. We cannot have [latex]c_2=0=-2[\/latex] so this boundary value problem has no solution.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"
Initial-Value Problems and Boundary-Value Problems<\/h2>\n
So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time.<\/p>\n
For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[\/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[\/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[\/latex]. With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions.<\/p>\n
This process is known as solving an initial-value problem<\/strong>. Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.<\/p>\n\ninitial-value problem<\/h3>\n
An initial-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at a single point in time [latex]t_0[\/latex]. For second-order equations, we need two initial conditions, typically [latex]y(t_0)[\/latex] and [latex]y'(t_0)[\/latex].<\/p>\n<\/section>\nSometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[\/latex] and [latex]y(t_1)=y_1[\/latex]. These conditions are called boundary conditions<\/strong>, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem<\/strong>.<\/p>\n\nboundary-value problem<\/h3>\n
A boundary-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at two different points in time, [latex]t_0[\/latex] and [latex]t_1[\/latex].<\/p>\n<\/section>\nWithin the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.<\/section>\nSolve the following initial-value problem: [latex]y''+3y'-4y=0[\/latex],\u00a0[latex]y(0)=1[\/latex],\u00a0[latex]y'(0)=-9[\/latex].<\/p>\n
boundary-value problem<\/h3>\r\nA boundary-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at two different points in time, [latex]t_0[\/latex] and [latex]t_1[\/latex].\r\n\r\n<\/section>Within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.<\/section>Solve the following initial-value problem: [latex]y''+3y'-4y=0[\/latex],\u00a0[latex]y(0)=1[\/latex],\u00a0[latex]y'(0)=-9[\/latex].[reveal-answer q=\"380210266\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380210266\"]\r\nWe already solved this differential equation previously and found the general solution to be:<\/p>\r\n
[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n
Then:<\/p>\r\n
[latex]y^\\prime=-4c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n
When [latex]x=0[\/latex], we have [latex]y(0)=c_1+c_2[\/latex] and [latex]y^\\prime(0)=-4c_1+c_2[\/latex]. Applying the initial conditions, we have:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1+c_2&=1 \\\\\r\n-4c_1+c_2&=-9\r\n\\end{aligned}[\/latex].<\/p>\r\n
Then [latex]c_1=1-c_2[\/latex]. Substituting this expression into the second equation, we see that:<\/p>\r\n
[latex]\\begin{aligned}\r\n-4(1-c_2)+c_2&=-9 \\\\\r\n-4+4c_2+c_2&=-9 \\\\\r\n5c_2&=-5 \\\\\r\nc_2&=-1\r\n\\end{aligned}[\/latex].<\/p>\r\n
So, [latex]c_1=2[\/latex] and the solution to the initial-value problem is:<\/p>\r\n
[latex]y(x)=2e^{-4x}-e^x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nSolve the following initial-value problem and graph the solution:<\/p>\r\n
[latex]y^{\\prime\\prime}+6y^\\prime+13y=0, \\ y(0)=0, \\ y^\\prime(0)=2[\/latex].<\/p>\r\n[reveal-answer q=\"835401193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"835401193\"]\r\n
We already solved this differential equation previously and found the general solution to be:<\/p>\r\n
[latex]y(x)=e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\nThen:\r\n
<\/div>\r\n[latex]y^\\prime(x)=e^{-3x}(-2c_1\\sin2x+2c_2\\cos2x)-3e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\n
When [latex]x=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y^\\prime(0)=2c_2-3c_1[\/latex]. Applying the initial conditions, we obtain:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1&=0 \\\\\r\n-3c_1+2c_2&= 2\r\n\\end{aligned}[\/latex].<\/p>\r\n
Therefore, [latex]c_1=0[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is shown in the following graph.<\/p>\r\n
[latex]y=e^{-3x}\\sin2x[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_3273\" align=\"aligncenter\" width=\"723\"] Figure 1.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nThe following initial-value problem models the position of an object with mass attached to a spring. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.<\/em><\/p>\r\n[latex]y^{\\prime\\prime}+2y^\\prime+y=0, \\ y(0)=1, \\ y^\\prime(0)=0[\/latex]<\/p>\r\n
Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[\/latex] sec? How fast is the mass moving at time [latex]t=1[\/latex] sec? In what direction?<\/p>\r\n[reveal-answer q=\"394824182\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394824182\"]\r\n
We already solved this previously and found the general solution to this differential equation to be:<\/p>\r\n
[latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex].<\/p>\r\n
Then:<\/p>\r\n
[latex]y^\\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[\/latex].<\/p>\r\n
When [latex]t=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y'(0)=-c_1+c_2[\/latex]. Applying the initial conditions, we obtain:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1&=1 \\\\\r\n-c_1+c_2&= 0\r\n\\end{aligned}[\/latex].<\/p>\r\n
Thus, [latex]c_1=1[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is:<\/p>\r\n
[latex]y(t)=e^{-t}+te^{-t}[\/latex].<\/p>\r\n
This solution is represented in the following graph. At time [latex]t=2[\/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\\approx0.406[\/latex] [latex]m[\/latex] below equilibrium.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3276\" align=\"aligncenter\" width=\"702\"] Figure 3.[\/caption]\r\n\r\n \r\n
To calculate the velocity at time [latex]t=1[\/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[\/latex], so:<\/p>\r\n
[latex]y^\\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[\/latex].<\/p>\r\n
Then [latex]y^\\prime(1)=e^{-1}\\approx-0.3679[\/latex]. At time [latex]t=1[\/latex], the mass is moving upward at [latex]0.3679[\/latex] m\/sec.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nEarlier we solved the differential equation [latex]y''+16y=0[\/latex] and found the general solution to be [latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:<\/p>\r\n\r\n
\r\n \t- [latex]y(0)=0[\/latex],\u00a0[latex]y\\left(\\frac{\\pi}4\\right)=0[\/latex]<\/li>\r\n \t
- [latex]y(0)=1[\/latex], [latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex]<\/li>\r\n \t
- [latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex],\u00a0[latex]y\\left(\\frac{3\\pi}8\\right)=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"437288237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437288237\"]\r\n
We have:<\/p>\r\n\r\n
\r\n[latex]y(x)=c_1\\cos{4t}+c_2\\sin{4t}[\/latex]<\/p>\r\n\r\n<\/div>\r\n
\r\n \t- Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[\/latex]. So the solution is of the form [latex]y(t)=c_2\\sin4t[\/latex]. When we apply the second boundary condition, though, we get [latex]y\\left(\\frac{\\pi}4\\right)=c_2\\sin\\left(4\\left(\\frac{\\pi}4\\right)\\right)=c_2\\sin\\pi=0[\/latex] for all values of [latex]c_2[\/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[\/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\\sin4t[\/latex] is a solution for any value of [latex]c_2[\/latex].<\/li>\r\n \t
- Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[\/latex]. Applying the second boundary condition gives[latex]y(\\frac{\\pi}{8})=c_2=0[\/latex], so [latex]c_2=0[\/latex]. In this case, we have a unique solution: [latex]y(t)=\\cos 4t[\/latex].<\/li>\r\n \t
- Applying the first boundary condition given here, we get [latex]y\\left(\\frac{\\pi}8\\right)=c_2=0[\/latex]. However, applying the second boundary condition gives [latex]y\\left(\\frac{3\\pi}8\\right)=-c_2=2[\/latex], so [latex]c_2=-2[\/latex]. We cannot have [latex]c_2=0=-2[\/latex] so this boundary value problem has no solution.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"
Initial-Value Problems and Boundary-Value Problems<\/h2>\n
So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time.<\/p>\n
For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[\/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[\/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[\/latex]. With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions.<\/p>\n
This process is known as solving an initial-value problem<\/strong>. Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.<\/p>\n\ninitial-value problem<\/h3>\n
An initial-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at a single point in time [latex]t_0[\/latex]. For second-order equations, we need two initial conditions, typically [latex]y(t_0)[\/latex] and [latex]y'(t_0)[\/latex].<\/p>\n<\/section>\nSometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[\/latex] and [latex]y(t_1)=y_1[\/latex]. These conditions are called boundary conditions<\/strong>, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem<\/strong>.<\/p>\n\nboundary-value problem<\/h3>\n
A boundary-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at two different points in time, [latex]t_0[\/latex] and [latex]t_1[\/latex].<\/p>\n<\/section>\nWithin the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.<\/section>\nSolve the following initial-value problem: [latex]y''+3y'-4y=0[\/latex],\u00a0[latex]y(0)=1[\/latex],\u00a0[latex]y'(0)=-9[\/latex].<\/p>\n
We already solved this differential equation previously and found the general solution to be:<\/p>\r\n
[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n
Then:<\/p>\r\n
[latex]y^\\prime=-4c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n
When [latex]x=0[\/latex], we have [latex]y(0)=c_1+c_2[\/latex] and [latex]y^\\prime(0)=-4c_1+c_2[\/latex]. Applying the initial conditions, we have:<\/p>\r\n
[latex]\\begin{aligned}\r\nc_1+c_2&=1 \\\\\r\n-4c_1+c_2&=-9\r\n\\end{aligned}[\/latex].<\/p>\r\n
Then [latex]c_1=1-c_2[\/latex]. Substituting this expression into the second equation, we see that:<\/p>\r\n
[latex]\\begin{aligned}\r\n-4(1-c_2)+c_2&=-9 \\\\\r\n-4+4c_2+c_2&=-9 \\\\\r\n5c_2&=-5 \\\\\r\nc_2&=-1\r\n\\end{aligned}[\/latex].<\/p>\r\n
So, [latex]c_1=2[\/latex] and the solution to the initial-value problem is:<\/p>\r\n
[latex]y(x)=2e^{-4x}-e^x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section> Solve the following initial-value problem and graph the solution:<\/p>\r\n [latex]y^{\\prime\\prime}+6y^\\prime+13y=0, \\ y(0)=0, \\ y^\\prime(0)=2[\/latex].<\/p>\r\n[reveal-answer q=\"835401193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"835401193\"]\r\n We already solved this differential equation previously and found the general solution to be:<\/p>\r\n [latex]y(x)=e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\nThen:\r\n [latex]y^\\prime(x)=e^{-3x}(-2c_1\\sin2x+2c_2\\cos2x)-3e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\n When [latex]x=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y^\\prime(0)=2c_2-3c_1[\/latex]. Applying the initial conditions, we obtain:<\/p>\r\n [latex]\\begin{aligned}\r\nc_1&=0 \\\\\r\n-3c_1+2c_2&= 2\r\n\\end{aligned}[\/latex].<\/p>\r\n Therefore, [latex]c_1=0[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is shown in the following graph.<\/p>\r\n [latex]y=e^{-3x}\\sin2x[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_3273\" align=\"aligncenter\" width=\"723\"] The following initial-value problem models the position of an object with mass attached to a spring. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.<\/em><\/p>\r\n [latex]y^{\\prime\\prime}+2y^\\prime+y=0, \\ y(0)=1, \\ y^\\prime(0)=0[\/latex]<\/p>\r\n Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[\/latex] sec? How fast is the mass moving at time [latex]t=1[\/latex] sec? In what direction?<\/p>\r\n[reveal-answer q=\"394824182\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394824182\"]\r\n We already solved this previously and found the general solution to this differential equation to be:<\/p>\r\n [latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex].<\/p>\r\n Then:<\/p>\r\n [latex]y^\\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[\/latex].<\/p>\r\n When [latex]t=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y'(0)=-c_1+c_2[\/latex]. Applying the initial conditions, we obtain:<\/p>\r\n [latex]\\begin{aligned}\r\nc_1&=1 \\\\\r\n-c_1+c_2&= 0\r\n\\end{aligned}[\/latex].<\/p>\r\n Thus, [latex]c_1=1[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is:<\/p>\r\n [latex]y(t)=e^{-t}+te^{-t}[\/latex].<\/p>\r\n This solution is represented in the following graph. At time [latex]t=2[\/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\\approx0.406[\/latex] [latex]m[\/latex] below equilibrium.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3276\" align=\"aligncenter\" width=\"702\"] To calculate the velocity at time [latex]t=1[\/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[\/latex], so:<\/p>\r\n [latex]y^\\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[\/latex].<\/p>\r\n Then [latex]y^\\prime(1)=e^{-1}\\approx-0.3679[\/latex]. At time [latex]t=1[\/latex], the mass is moving upward at [latex]0.3679[\/latex] m\/sec.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section> Earlier we solved the differential equation [latex]y''+16y=0[\/latex] and found the general solution to be [latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:<\/p>\r\n\r\n We have:<\/p>\r\n\r\n [latex]y(x)=c_1\\cos{4t}+c_2\\sin{4t}[\/latex]<\/p>\r\n\r\n<\/div>\r\n So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time.<\/p>\n For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[\/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[\/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[\/latex]. With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions.<\/p>\n This process is known as solving an initial-value problem<\/strong>. Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.<\/p>\n An initial-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at a single point in time [latex]t_0[\/latex]. For second-order equations, we need two initial conditions, typically [latex]y(t_0)[\/latex] and [latex]y'(t_0)[\/latex].<\/p>\n<\/section>\n Sometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[\/latex] and [latex]y(t_1)=y_1[\/latex]. These conditions are called boundary conditions<\/strong>, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem<\/strong>.<\/p>\n A boundary-value problem<\/strong> requires finding a solution to a differential equation that satisfies specific conditions at two different points in time, [latex]t_0[\/latex] and [latex]t_1[\/latex].<\/p>\n<\/section>\n Figure 1.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section> Figure 3.[\/caption]\r\n\r\n \r\n\r\n \t
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Initial-Value Problems and Boundary-Value Problems<\/h2>\n
initial-value problem<\/h3>\n
boundary-value problem<\/h3>\n