Unit 1 - Practice Quiz

The necessary and sufficient condition for a first-order differential equation to be exact is that the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x.

By comparing the given equation with the standard form , the function multiplying is and the function multiplying is . Thus, and .

An equation is exact if there exists a function such that its total differential is equal to . This implies and .

Here, and . We have and . Since , the equation is not exact.

An integrating factor, often denoted by , is a function that is used to transform a non-exact differential equation of the form into an exact one.

This is a standard rule for finding an integrating factor. If , then the integrating factor is given by the formula I.F. .

A non-exact equation cannot be solved directly using the integration method for exact equations. The integrating factor is a tool to convert it into an exact form, which can then be easily solved.

This is another standard rule for finding an integrating factor. If , then the integrating factor is given by the formula I.F. .

For notational convenience in equations of first order and higher degree, is used as a shorthand for the first derivative, .

The degree of a differential equation is the highest power of the highest order derivative, after the equation has been cleared of radicals and fractions. Here, the highest derivative is and its highest power is 3.

This is a quadratic equation in . It can be factored into , which gives two distinct linear differential equations for . This is the characteristic of an equation 'solvable for p'.

An equation is of the first order if the highest derivative is . It is of a 'higher degree' if this derivative appears with a power (degree) of 2 or more.

Clairaut's equation is a special type of first-order, higher-degree differential equation defined by the form , where .

A remarkable property of Clairaut's equation is that its general solution is found simply by replacing the derivative with an arbitrary constant , giving the family of straight lines .

The equation is in the slope-intercept form where the slope and the y-intercept . For each value of the constant , this represents a straight line.

This equation matches the general form , where . The other options do not fit this specific structure.

The singular solution of a Clairaut's equation is the envelope of the family of straight lines given by the general solution. It cannot be obtained by assigning a specific value to the arbitrary constant.

The solution of an exact equation is of the form . For different values of the constant , this equation defines the level curves of the surface .

If an equation of degree 'n' in p is solvable for p, it can be broken down into 'n' distinct first-order, first-degree equations of the form . Each of these can have a solution passing through a given point.

To find the general solution of any Clairaut's equation, we simply replace the term (which represents ) with an arbitrary constant . Here, , so the solution is .

For an equation to be exact, we must have . Here, and . Differentiating, we get and . Equating these two expressions gives , which simplifies to . Therefore, .

Let and . We find and . Since , the equation is exact. The solution is . Using the initial condition , we get . Thus, the solution is .

The equation is exact since and . To find the solution , we integrate with respect to : . Differentiating this with respect to gives , which must equal . So, , which implies . Integrating gives . The solution is .

Let and . We compute and . The equation is not exact. We check the expression . Since this is a function of alone, the integrating factor (I.F.) is .

The equation is of the form . Here and . An integrating factor for this form is given by . We calculate . So, a valid integrating factor is . Since any constant multiple of an integrating factor is also an integrating factor, is a valid choice.

Let and . We have and . The equation is not exact. We check the expression . Since this is a function of alone, the integrating factor (I.F.) is .

The integrating factor is . Multiplying the equation by it gives . Let this be . Now and , so it is exact. The solution is . Multiplying by gives .

Let . The equation becomes . Factoring the quadratic, we get . This gives two possibilities: or . Case 1: , or . Case 2: , or . The general solution is the product of these two solutions with a common constant : .

The equation is cubic in , so it's not easily solvable for . It is also not linear in . It is not in the Clairaut's form . However, we can rearrange the equation to isolate : . This expresses as a function of and , making it an equation of the form , which is solvable for by differentiating with respect to .

This is a quadratic equation in . Using the quadratic formula, . This gives two solutions for : and . Integrating these gives two families of solutions: and . The complete solution is .

The equation is a quadratic in . We can factor it: . This gives two possibilities. First, , which is the given solution. Second, . This is a separable equation: . Integrating both sides gives .

A Clairaut's equation has the form . Its general solution is found by replacing the parameter with an arbitrary constant . In this case, . Therefore, the general solution is .

The general solution is . To find the singular solution (the envelope of the family of lines), we differentiate the general solution with respect to and set it to zero: . From this, . Substituting back into the general solution: . Squaring both sides gives the equation of the envelope: .

The general solution is a family of straight lines . To find the singular solution, we differentiate with respect to : . This gives . From the general solution, we have . From the derivative equation, , and from the second equation, . We can also use the parametric form of the envelope: and . Here , so . This gives and . Squaring and adding these two equations gives . This is the equation of a unit circle.

This is a Clairaut's equation. The general solution is , valid for . Differentiating with respect to to find the envelope gives , so . For , we must have . Substituting this back into the general solution gives the singular solution: .

First, verify exactness. Let and . Then and . The equation is exact. The solution is found by integrating . Given the point , we substitute : .

This question tests the standard rules for finding integrating factors. If , the I.F. is . If , the I.F. is . This is a direct application of the formula.

The equation is of the form . Let and . The integrating factor for this type is . We calculate . The integrating factor is . Since a constant multiple does not affect its property as an integrating factor, is a correct answer.

For equations of the form , the standard method of solution is to differentiate the entire equation with respect to . This gives . This results in a new differential equation involving , , and , which is often solvable for as a function of .

First, confirm exactness: and . They are equal. The solution is . . Differentiating with respect to gives . Thus, , so . The general solution is . Substituting the point : . The constant is 5.

Let the integrating factor be where . The equation must be exact. The condition for exactness is .
Using the product rule: .
Since is a function of , we have and .
Substituting these in: .
Rearranging terms: .
This gives .
Since the left side, , is a function of , the right side must also be a function of for an integrating factor of this form to exist.

Given . Let . Differentiating with respect to , we get . Let , so .
Substitute into the original equation: .
Multiply by : .
Since , we get . This is a Clairaut's equation for .
To find the singular solution, we eliminate from this equation and its derivative with respect to : .
From the second equation, .
Substitute this into the Clairaut's form: .
So .
Substitute : .
Finally, substitute back : .

The equation is .
p-discriminant: This is a quadratic in . The condition for repeated roots gives the loci. Setting the coefficients of powers of or the constant term to zero, we get and . So the p-discriminant is .

General Solution: We solve for : . Integrating gives .

c-discriminant: We eliminate from the general solution and its derivative with respect to : . Substituting this into the solution gives , which implies or . The c-discriminant is .

Analysis:

• The factor is common to both discriminants and appears with power 1, indicating is the envelope (singular solution).
• The factor appears only in the p-discriminant, indicating is a tac-locus.
• The factor appears only in the c-discriminant, indicating is a node-locus.

For the equation to be exact, we must have .
Here, .
And .

First, calculate :


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Next, calculate :


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Equating the two partial derivatives:
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For this to hold true for all , we must have .

Given and . We check for an integrating factor . The formula requires calculating .
, .
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So, . This is a function of .
The integrating factor is found by solving .
. So, .

Multiply the original equation by :

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This equation is exact. Let and .
The solution is found by integrating:
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To find , we differentiate with respect to : .
Thus, , which gives .
The solution is , which simplifies to .

Let and .
For exactness, .
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Now that , the equation is .
Integrate with respect to : .
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So , which means .

First, multiply the equation by the integrating factor :
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Let the new components be and .
Let's verify it is exact:
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Since , the equation is exact.
To find the solution , we can integrate with respect to , as it's simpler:
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Now, differentiate with respect to and set it equal to :
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This simplifies to .
We need to integrate to find . This can be done by parts or by noticing that .
So, .
The general solution is , which can be written as .

The given equation is in Clairaut's form, .
The general solution is a family of straight lines given by .
The singular solution is the envelope of this family. To find it, we eliminate from the given equation and its derivative with respect to :
1)
2) Differentiating with respect to :
From (2), we get .
Substitute this expression for back into (1):
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Now we have parametric equations for the envelope in terms of :
and .
To get the Cartesian equation, we can eliminate . From the expressions for and , we can write:
and .
Squaring both equations:
and .
Adding them together: .
This is the equation , which represents an ellipse.

The equation is given in the form , which is solvable for . We differentiate with respect to to solve it.
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Differentiating with respect to : .
We know that .
So, .
Rearranging to solve for : .
Now, we integrate both sides to find :
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The first integral is .
The second integral, , requires integration by parts ().
Let and . Then and .
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Combining the results, we get:
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Let the integrating factor be . Multiplying the DE by gives:
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Let and .
For the new equation to be exact, we need .
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Equating the two partial derivatives:
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Dividing by (for ):
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Alternatively, for an integrating factor that is a function of alone, the expression must be a function of .
. .
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The integrating factor is . Thus, .

The relationship between the potential function and the components and of an exact equation is given by . Therefore, we must have and .
Given .

We calculate the partial derivative with respect to to find :
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Treating as a constant: .

We calculate the partial derivative with respect to to find :
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Using the chain rule for the first term and standard derivative for the second:
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So, the pair is .

The given equation is , which is solvable for . We differentiate the entire equation with respect to :
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This equation is separable if we rearrange it for : .
This gives a linear first-order ODE for as a function of : .
The integrating factor is .
Multiplying the linear DE by : .
The left side is the derivative of a product: .
Integrating with respect to : .
Solving for : .
Now, substitute this expression for back into the original equation :
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The general solution is given in parametric form with parameter : and .

The given equation is . Taking the square root of both sides gives two separate equations:
1) .
2) .
Both are in Clairaut's form . They share the same general solution family, . The singular solution will be the envelope of this family.
Let's find the singular solution for the first equation, .
Differentiate with respect to : .
This gives .
Substitute back into the equation for : .
We have and .
To eliminate , notice that .
So, the singular solution is . Following the same procedure for the second equation yields the exact same singular solution. The shape is a hyperbola.

The equation is . This equation is solvable for . Let's rewrite it as . This is a form of Lagrange's equation. Let's try to find the p-discriminant. Consider the equation as a cubic in : . The singular solution is found by eliminating between and .
, which gives .
From the original equation, .
We have , so .
Let's substitute from the derivative into the original equation: .
Now we have two relations for : and .
Cube the first relation: .
Square the second relation: .
Equating the expressions for : .
Assuming $y
eq 016y^3\frac{4x^3}{27} = y$.
Thus, the singular solution is . The solution is also obtained, and it is a particular solution.

Let the integrating factor be . The new equation is .


For exactness, .


Equating the coefficients of like terms:
For the term: .
For the term: .
We have a system of two linear equations:
1)
2)
Subtracting (2) from (1): .
Substitute into (2): .
The integrating factor is .
The question asks for the value of , which is directly given by the second equation: .

The general solution is given as . This is a family of curves parameterized by . The singular solution is the envelope of this family. To find the envelope, we use the c-discriminant method. We need to eliminate the parameter from the general solution and its partial derivative with respect to .
1) The general solution: .
2) Differentiate with respect to : .
From equation (2), we get , which implies .
Substitute back into the general solution (1):
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So, the singular solution is . Geometrically, the family of parabolas all have their vertices on the x-axis and are tangent to it. The x-axis () is therefore the envelope of this family.

The question should be: Find the solution of . First, check for exactness. , . . . It is not exact. Let's find an integrating factor. Calculate . Since this is a function of alone, the integrating factor is . Multiply the original equation by : . This new equation is exact. Let and . The solution is found by integration. . Differentiate this with respect to to find : . This implies , so . The general solution is .

The given equation is . This can be factored as . This is not a Clairaut's equation. To solve it, we can differentiate with respect to . Let's try taking the square root: (assuming ). Then . So . This is a non-linear first-order ODE. Let's try another way. Differentiate the original equation w.r.t : . . . This looks complicated. Let's re-examine . Differentiating wrt x: . This means either or . Case 1: . Then . So is a solution. Let's check: if , then . . So is not a general solution. Case 2: From , this is too complex. Notice that is a potential solution. Let's check. . We need to see if is satisfied. Substitute into . . So must be the solution. . For $p

The equation is , where . We can solve for : . This gives two first-order separable equations.
Case 1: .
Separating variables: .
Integrating both sides: .
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Case 2: .
Separating variables: .
Integrating both sides: .
. Let , then .
The family covers both cases, as the constant can incorporate the sign change.
To find singular solutions, we examine the p-discriminant of the original equation . The discriminant is . Setting this to zero for repeated roots gives , so and .
Let's check if they are solutions. If , then . Substituting into the DE: , which is true. If , then . Substituting: , which is also true. These constant functions are the horizontal envelopes for the family of sine curves . Hence, they are singular solutions.

For the equation to be exact, we must have .
Given and , the exactness condition becomes:
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To solve this, we can separate the variables and :
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The left side is a function of only, and the right side is a function of only. For them to be equal for all and , they must both be equal to the same constant, say .
So, and .
This implies and .
The solution is . We have . And .

Let's rewrite the equation by expanding the terms:


This equation is of the form . For this form, the integrating factor is .
Let's calculate :


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The integrating factor is . Since a constant factor does not matter, we can take .
Let's check the question's premise. Maybe the form is not general. Let's use the formula for .




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Let . The integrating factor satisfies .
. So . So .
There seems to be a discrepancy. Let's re-read the original equation. It's . Let's try .
New M: .
New N: .
New M: . .
New N: . .