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Differential Equations : Initial-Value Problems

Study concepts, example questions & explanations for differential equations, all differential equations resources, example questions, example question #1 : initial value problems.

$initial value problem number of solutions$

First identify what is known.

The general function is,

$initial value problem number of solutions$

The initial value is six in mathematical terms is,

$initial value problem number of solutions$

So this is a separable differential equation, but it is also subject to an initial condition. This means that you have enough information so that there should not be a constant in the final answer.

You start off by getting all of the like terms on their respective sides, and then taking the anti-derivative. Your pre anti-derivative equation will look like:

$initial value problem number of solutions$

Then taking the anti-derivative, you include a C value:

$initial value problem number of solutions$

Then, using the initial condition given, we can solve for the value of C:

$initial value problem number of solutions$

Solving for C, we get

$initial value problem number of solutions$

Example Question #3 : Initial Value Problems

$initial value problem number of solutions$

Example Question #4 : Initial Value Problems

$initial value problem number of solutions$

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

$initial value problem number of solutions$

Then integrate, and make sure to add a constant at the end

$initial value problem number of solutions$

Plug in the initial condition to get:

$initial value problem number of solutions$

Solve the separable differential equation

$initial value problem number of solutions$

none of these answers

$initial value problem number of solutions$

To start off, gather all of the like variables on separate sides.

Notice that when you divide sec(y) to the other side, it will just be cos(y),

and the csc(x) on the bottom is equal to sin(x) on the top.

$initial value problem number of solutions$

In order to solve for y, we just need to take the arcsin of both sides:

$initial value problem number of solutions$

Solve the differential equation

$initial value problem number of solutions$

Then, after the anti-derivative, make sure to add the constant C:

$initial value problem number of solutions$

Solve for y

$initial value problem number of solutions$

None of these answers

$initial value problem number of solutions$

Taking the anti-derivative once, we get:

$initial value problem number of solutions$

we get the final answer of:

$initial value problem number of solutions$

Example Question #8 : Initial Value Problems

Solve the differential equation for y

$initial value problem number of solutions$

subject to the initial condition:

$initial value problem number of solutions$

Solving for C:

$initial value problem number of solutions$

Then taking the square root to solve for y, we get:

$initial value problem number of solutions$

Example Question #2 : Initial Value Problems

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no of solutions of the initial value problem?

$x \dfrac{dy}{dx} = y , y (0) = 0, x \geq 0 .$

My Approach :

$\dfrac{dy}{y} = \dfrac{dx}{x},$ by variable separable method, we get

$lny = ln x +c $ and then raising e to both sides will get $ y=x+c.$ Then substituting the intial condtions, we get $c=-1$

so $y = x -1 $is the required solution.

But it's given that it has uncountable number of solution. I don't understand it.

2 $\begingroup$ You've not solved the differential equation correctly. You will get $\ln|y|=\ln|x|+C$ or $|y|=K|x|$ where $K=e^C$ $\endgroup$ – OnceUponACrinoid Feb 7, 2016 at 15:30

2 Answers 2

If you solve the differential equation correctly, then $$\ln |y|= \ln x + c = \ln e^c x,$$ and so $$y=\pm e^c x,$$ not $y=x+c$. Then $e^c \cdot 0 = 0$ for all constant $C$, given initial value doesn't give us a useful information.

$$x\cdot\frac{\text{d}y(x)}{\text{d}x}=y(x)\Longleftrightarrow$$ $$xy'(x)=y(x)\Longleftrightarrow$$ $$y'(x)=\frac{y(x)}{x}\Longleftrightarrow$$ $$\frac{y'(x)}{y(x)}=\frac{1}{x}\Longleftrightarrow$$ $$\int\frac{y'(x)}{y(x)}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$ $$\ln|y(x)|=\ln|x|+\text{C}\Longleftrightarrow$$ $$y(x)=e^{\ln|x|+\text{C}}\Longleftrightarrow$$ $$y(x)=\text{C}e^{\ln|x|}\Longleftrightarrow$$ $$y(x)=\text{C}x$$

The given initial condition is not sufficient to solve for the unknown constant, because:

If, $y(0)=0$:

$$0=\text{C}\cdot0\Longleftrightarrow \text{C}=\frac{0}{0}$$

And we can't divide by $0$.

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Module 4: Differential Equations

Initial-value problems, learning outcomes.

Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value , which is an ordered pair that is used to find a particular solution.

A differential equation together with one or more initial values is called an initial-value problem . The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation [latex]{y}^{\prime }=2x[/latex], then [latex]y\left(3\right)=7[/latex] is an initial value, and when taken together, these equations form an initial-value problem. The differential equation [latex]y\text{''}-3{y}^{\prime }+2y=4{e}^{x}[/latex] is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be [latex]y\left(0\right)=2[/latex] and [latex]{y}^{\prime }\left(0\right)=-1[/latex]. These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is [latex]t[/latex], which represents time. Thus, a value of [latex]t=0[/latex] represents the beginning of the problem.

Example: Verifying a Solution to an Initial-Value Problem

Verify that the function [latex]y=2{e}^{-2t}+{e}^{t}[/latex] is a solution to the initial-value problem

For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that [latex]y[/latex] satisfies the differential equation, we start by calculating [latex]{y}^{\prime }[/latex]. This gives [latex]{y}^{\prime }=-4{e}^{-2t}+{e}^{t}[/latex]. Next we substitute both [latex]y[/latex] and [latex]{y}^{\prime }[/latex] into the left-hand side of the differential equation and simplify:

This is equal to the right-hand side of the differential equation, so [latex]y=2{e}^{-2t}+{e}^{t}[/latex] solves the differential equation. Next we calculate [latex]y\left(0\right)\text{:}[/latex]

This result verifies the initial value. Therefore the given function satisfies the initial-value problem.

Watch the following video to see the worked solution to Example: Verifying a Solution to an Initial-Value Problem

You can view the transcript for this segmented clip of “4.1.5” here (opens in new window) .

Verify that [latex]y=3{e}^{2t}+4\sin{t}[/latex] is a solution to the initial-value problem

First verify that [latex]y[/latex] solves the differential equation. Then check the initial value.

In the previous example, the initial-value problem consisted of two parts. The first part was the differential equation [latex]{y}^{\prime }+2y=3{e}^{x}[/latex], and the second part was the initial value [latex]y\left(0\right)=3[/latex]. These two equations together formed the initial-value problem.

The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of [latex]C[/latex]. The family of solutions to the differential equation in the example is given by [latex]y=2{e}^{-2t}+C{e}^{t}[/latex]. This family of solutions is shown in Figure 2, with the particular solution [latex]y=2{e}^{-2t}+{e}^{t}[/latex] labeled.

A graph of a family of solutions to the differential equation y’ + 2 y = 3 e ^ t, which are of the form y = 2 e ^ (-2 t) + C e ^ t. The versions with C = 1, 0.5, and -0.2 are shown, among others not labeled. For all values of C, the function increases rapidly for [latex]t[/latex] < 0 as [latex]t[/latex] goes to negative infinity. For C > 0, the function changes direction and increases in a gentle curve as [latex]t[/latex] goes to infinity. Larger values of C have a tighter curve closer to the [latex]y[/latex]-axis and at a higher y value. For C = 0, the function goes to 0 as [latex]t[/latex] goes to infinity. For C < 0, the function continues to decrease as [latex]t[/latex] goes to infinity.

Figure 2. A family of solutions to the differential equation [latex]{y}^{\prime }+2y=3{e}^{t}[/latex]. The particular solution [latex]y=2{e}^{-2t}+{e}^{t}[/latex] is labeled.

Example: Solving an Initial-value Problem

Solve the following initial-value problem:

The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation

We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex]. Solving the previous equation for [latex]y[/latex] gives

Because [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex] are both constants, [latex]{C}_{2}-{C}_{1}[/latex] is also a constant. We can therefore define [latex]C={C}_{2}-{C}_{1}[/latex], which leads to the equation

Next we determine the value of [latex]C[/latex]. To do this, we substitute [latex]x=0[/latex] and [latex]y=5[/latex] into our aforementioned equation and solve for [latex]C\text{:}[/latex]

Now we substitute the value [latex]C=2[/latex] into our equation. The solution to the initial-value problem is [latex]y=3{e}^{x}+\frac{1}{3}{x}^{3}-4x+2[/latex].

The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.

Solve the initial-value problem

First take the antiderivative of both sides of the differential equation. Then substitute [latex]x=0[/latex] and [latex]y=8[/latex] into the resulting equation and solve for [latex]C[/latex].

[latex]y=\frac{1}{3}{x}^{3}-2{x}^{2}+3x - 6{e}^{x}+14[/latex]

In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth’s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with Newton’s second law of motion (in equation form [latex]F=ma[/latex], where [latex]F[/latex] represents force, [latex]m[/latex] represents mass, and [latex]a[/latex] represents acceleration), to derive an equation that can be solved.

A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m/sec ^ 2.

Figure 3. For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).

In Figure 3. we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth’s surface, [latex]g[/latex], is approximately [latex]9.8{\text{m/s}}^{2}[/latex]. We introduce a frame of reference, where Earth’s surface is at a height of 0 meters. Let [latex]v\left(t\right)[/latex] represent the velocity of the object in meters per second. If [latex]v\left(t\right)>0[/latex], the ball is rising, and if [latex]v\left(t\right)<0[/latex], the ball is falling (Figure 4).

A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) > 0, and the arrow pointing down is labeled v(t) < 0.

Figure 4. Possible velocities for the rising/falling baseball.

Our goal is to solve for the velocity [latex]v\left(t\right)[/latex] at any time [latex]t[/latex]. To do this, we set up an initial-value problem. Suppose the mass of the ball is [latex]m[/latex], where [latex]m[/latex] is measured in kilograms. We use Newton’s second law, which states that the force acting on an object is equal to its mass times its acceleration [latex]\left(F=ma\right)[/latex]. Acceleration is the derivative of velocity, so [latex]a\left(t\right)={v}^{\prime }\left(t\right)[/latex]. Therefore the force acting on the baseball is given by [latex]F=m{v}^{\prime }\left(t\right)[/latex]. However, this force must be equal to the force of gravity acting on the object, which (again using Newton’s second law) is given by [latex]{F}_{g}=\text{-}mg[/latex], since this force acts in a downward direction. Therefore we obtain the equation [latex]F={F}_{g}[/latex], which becomes [latex]m{v}^{\prime }\left(t\right)=\text{-}mg[/latex]. Dividing both sides of the equation by [latex]m[/latex] gives the equation

Notice that this differential equation remains the same regardless of the mass of the object.

We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the initial velocity , or the velocity at time [latex]t=0[/latex]. This is denoted by [latex]v\left(0\right)={v}_{0}[/latex].

Example: Velocity of a Moving Baseball

A baseball is thrown upward from a height of [latex]3[/latex] meters above Earth’s surface with an initial velocity of [latex]10\text{m/s}[/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15\text{kg}[/latex] at Earth’s surface.

where [latex]g=9.8{\text{m/s}}^{2}[/latex]. The initial condition is [latex]v\left(0\right)={v}_{0}[/latex], where [latex]{v}_{0}=10\text{m/s}\text{.}[/latex] Therefore the initial-value problem is [latex]{v}^{\prime }\left(t\right)=-9.8{\text{m/s}}^{2},v\left(0\right)=10\text{m/s}\text{.}[/latex] The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives

The next step is to solve for [latex]C[/latex]. To do this, substitute [latex]t=0[/latex] and [latex]v\left(0\right)=10\text{:}[/latex]

Suppose a rock falls from rest from a height of [latex]100[/latex] meters and the only force acting on it is gravity. Find an equation for the velocity [latex]v\left(t\right)[/latex] as a function of time, measured in meters per second.

What is the initial velocity of the rock? Use this with the differential equation in the example: Velocity of a Moving Baseball to form an initial-value problem, then solve for [latex]v\left(t\right)[/latex].

[latex]v\left(t\right)=-9.8t[/latex]

A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a given point in time. Let [latex]s\left(t\right)[/latex] denote the height above Earth’s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation [latex]{s}^{\prime }\left(t\right)=v\left(t\right)[/latex]. An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation [latex]s\left(0\right)={s}_{0}[/latex]. Together these assumptions give the initial-value problem

If the velocity function is known, then it is possible to solve for the position function as well.

Example: Height of a Moving Baseball

Next we substitute [latex]t=0[/latex] and solve for [latex]C\text{:}[/latex]

How to solve initial value problems

What is an initial value problem?

Consider the following situation. You’re given the function ???f(x)=2x-3??? and asked to find its derivative.

This function is pretty basic, so unless you’re taking calculus out of order, it shouldn’t cause you too much stress to figure out that the derivative of ???f(x)??? is ???2???.

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Now consider what it would be like to work backwards from our derivative. If you’re given the function ???f'(x)=2??? and asked to find its integral, it’s impossible for you to get back to the original function, ???f(x)=2x-3???. As you can see, taking the integral of the derivative we found gives us back the first term of the original function, ???2x???, but somewhere along the way we lost the ???-3???. In fact, we always lose the constant (term without a variable attached), when we take the derivative of something. Which means we’re never going to get the constant back when we try to integrate our derivative. It’s lost forever.

Accounting for that lost constant is why we always add ???C??? to the end of our integrals. ???C??? is called the “constant of integration” and it acts as a placeholder for our missing constant. In order to get back to our original function, and find our long-lost friend, ???-3???, we’ll need some additional information about this problem, namely, an initial condition, which looks like this:

???y(0)=-3???

Problems that provide you with one or more initial conditions are called Initial Value Problems. Initial conditions take what would otherwise be an entire rainbow of possible solutions, and whittles them down to one specific solution.

Remember that the basic idea behind Initial Value Problems is that, once you differentiate a function, you lose some information about that function. More specifically, you lose the constant. By integrating ???f'(x)???, you get a family of solutions that only differ by a constant.

???\int 2\ dx=2x-3???

???\int 2\ dx=2x+7???

???\int 2\ dx=2x-\sqrt{2}???

Given one point on the function, (the initial condition), you can pick a specific solution out of a much broader solution set.

How to solve initial value problems by using the initial condition

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Want to learn more about calculus 2 i have a step-by-step course for that. :), find the antiderivative, then evaluate at the initial condition.

Given ???f'(x)=2??? and ???f(0)=-3???, find ???f(x)???.

Integrating ???f'(x)??? means we’re integrating ???2\ dx???, and we’ll get ???2x+C???, where ???C??? is the constant of integration. At this point, ???C??? is holding the place of our now familiar friend, ???-3???, but we don’t know that yet. We have to use our initial condition to find out.

Initial value problems for Calculus 2.jpg

Notice that the solution would have been different had we been given a different initial condition.

To use our initial condition, ???f(0)=-3???, we plug in the number inside the parentheses for ???x??? and the number on the right side of the equation for ???y???. Therefore, in our case, we’ll plug in ???0??? for ???x??? and ???-3??? for ???y???.

???-3=2(0)+C???

Notice that the solution would have been different had we been given a different initial condition. Now we know exactly what the full solution looks like, and exactly which one of the many possible solutions was originally differentiated. Therefore, the final answer is the function we originally differentiated:

???f(x)=2x-3???

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