Question #19

Practice Question

A.$y = e^{-x} + Ce^{-2x}$
B.$y = e^{-2x} + Ce^{-x}$
C.$y = -e^{-x} + Ce^{-2x}$
D.$y = -e^{-2x} + Ce^{-x}$

Concept Applied

Integrating factor is $e^{2x}$; multiply and integrate to get solution....

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Question #20

Practice Question

Concept Applied

Separate variables: $\frac{dy}{y} = \frac{dx}{x}$, integrate: $\ln y = \ln x + C$, apply IC: $y = x$, so $y(e) = e \approx 2.72$....

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.$x + y + z = 3$
B.$x^2 + y^2 + z^2 = 3$
C.$xy + yz + zx = 1$
D.$x^2 + y^2 + z^2 = 1$

Concept Applied

Step 1: Since the surface is passing through the point $(1, 1, 1)$, we can substitute these values into the equation of the surface to get

Question #22

Practice Question

A.$y = Ae^{x^2}$
B.$y = Ae^{-x^2}$
C.$y = Ae^{x^2} + B$
D.$y = Ae^{-x^2} + B$

Concept Applied

Step 1: Given the differential equation $\frac{dy}{dx} = \frac{2x}{y}$, we can rewrite it as $\frac{dy}{y} = \frac{2x}{y}dx$....

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Question #23

Practice Question

A.$v = y^{-1}$
B.$v = y^2$
C.$v = \ln y$
D.$v = e^y$

Concept Applied

Bernoulli equation with $n=2$, use $v = y^{1-2} = y^{-1}$....

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Question #24

Practice Question

A.$y = x$
B.$y = 2x$
C.$y = \frac{1}{x}$
D.$y = x^2$

Concept Applied

Separate variables: $\frac{dy}{y} = \frac{dx}{x}$, integrate and apply $y(1)=1$....

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.Second order, first degree
B.Second order, second degree
C.First order, second degree
D.First order, first degree

Concept Applied

The highest derivative is \(d^2y/dx^2\) → order 2. The equation contains \((dy/dx)^2\) which makes the degree 2 (since the highest power of the highes...

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Question #26

Practice Question

A.$x^2 + 3xy + 2y^2 = C$
B.
.$x^2 + 6xy + 4y^2 = C$
D.
...

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Question #27

Practice Question

A.$x^2$
B.
.$\frac{1}{x^2}$
D.$e^{2x}$

Concept Applied

Integrating factor is $e^{\int (2/x) dx} = e^{2\ln x} = x^2$....

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Question #28

Practice Question

A.-3
B.3
C.-\frac{1}{3}
D.\frac{1}{3}

Concept Applied

Separate variables: $\frac{dy}{y^{2}} = (x^{2}+1)dx$. Integrate: $-\frac{1}{y}=\frac{x^{3}}{3}+x+C$. Use $y(0)=1$ to get $C=-1$. Hence $-\frac{1}{y}=\...

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Question #29

Practice Question

A.$y = x$
B.$y = 2x$
C.$y = \frac{1}{x}$
D.$y = x^2$

Concept Applied

Separate variables and integrate: $\ln|y| = \ln|x| + C$, apply IC....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.$y = vx$
B.$x = vy$
C.$y = x^2v$
D.$y = \sqrt{v}$

Concept Applied

The given equation is homogeneous. Substituting $y = vx$ gives $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The right-hand side becomes $\frac{x^2 + (vx)^2}{...

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Question #31

Practice Question

A.Order 2, Degree 3
B.Order 2, Degree 4
C.Order 1, Degree 3
D.Order 3, Degree 2

Concept Applied

Highest derivative is $d^2y/dx^2$ (order 2); its power is 3 (degree 3)....

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Question #32

Practice Question

Concept Applied

Use $P(t) = P_0 e^{kt}$; solve

...

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Question #33

Practice Question

A.$y = e^{-x} + Ce^{-2x}$
B.$y = e^{-2x} + Ce^{-x}$
C.$y = -e^{-x} + Ce^{-2x}$
D.$y = -e^{-2x} + Ce^{-x}$

Concept Applied

Integrating factor is $e^{2x}$; multiply and integrate to get solution....

Read Full Step-by-Step Solution →

Question #34

Practice Question

Concept Applied

Separate variables: $\frac{dy}{y} = \frac{dx}{x}$, integrate: $\ln y = \ln x + C$, apply IC: $y = x$, so $y(e) = e \approx 2.72$....

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.$x + y + z = 3$
B.$x^2 + y^2 + z^2 = 3$
C.$xy + yz + zx = 1$
D.$x^2 + y^2 + z^2 = 1$

Concept Applied

Step 1: Since the surface is passing through the point $(1, 1, 1)$, we can substitute these values into the equation of the surface to get

Question #36

Practice Question

A.$y = Ae^{x^2}$
B.$y = Ae^{-x^2}$
C.$y = Ae^{x^2} + B$
D.$y = Ae^{-x^2} + B$

Concept Applied

Step 1: Given the differential equation $\frac{dy}{dx} = \frac{2x}{y}$, we can rewrite it as $\frac{dy}{y} = \frac{2x}{y}dx$....

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.$v = y^{-1}$
B.$v = y^2$
C.$v = \ln y$
D.$v = e^y$

Concept Applied

Bernoulli equation with $n=2$, use $v = y^{1-2} = y^{-1}$....

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.$y = x$
B.$y = 2x$
C.$y = \frac{1}{x}$
D.$y = x^2$

Concept Applied

Separate variables: $\frac{dy}{y} = \frac{dx}{x}$, integrate and apply $y(1)=1$....

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.Second order, first degree
B.Second order, second degree
C.First order, second degree
D.First order, first degree

Concept Applied

The highest derivative is \(d^2y/dx^2\) → order 2. The equation contains \((dy/dx)^2\) which makes the degree 2 (since the highest power of the highes...

Read Full Step-by-Step Solution →

Question #40

Practice Question

A.$x^2 + 3xy + 2y^2 = C$
B.
.$x^2 + 6xy + 4y^2 = C$
D.
...

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.$x^2$
B.
.$\frac{1}{x^2}$
D.$e^{2x}$

Concept Applied

Integrating factor is $e^{\int (2/x) dx} = e^{2\ln x} = x^2$....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.-3
B.3
C.-\frac{1}{3}
D.\frac{1}{3}

Concept Applied

Separate variables: $\frac{dy}{y^{2}} = (x^{2}+1)dx$. Integrate: $-\frac{1}{y}=\frac{x^{3}}{3}+x+C$. Use $y(0)=1$ to get $C=-1$. Hence $-\frac{1}{y}=\...

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.$y = x$
B.$y = 2x$
C.$y = \frac{1}{x}$
D.$y = x^2$

Concept Applied

Separate variables and integrate: $\ln|y| = \ln|x| + C$, apply IC....

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.$y = vx$
B.$x = vy$
C.$y = x^2v$
D.$y = \sqrt{v}$

Concept Applied

The given equation is homogeneous. Substituting $y = vx$ gives $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The right-hand side becomes $\frac{x^2 + (vx)^2}{...

Read Full Step-by-Step Solution →

Question #45

Practice Question

A.Order 2, Degree 3
B.Order 2, Degree 4
C.Order 1, Degree 3
D.Order 3, Degree 2

Concept Applied

Highest derivative is $d^2y/dx^2$ (order 2); its power is 3 (degree 3)....

Read Full Step-by-Step Solution →

Question #46

Practice Question

Concept Applied

Use $P(t) = P_0 e^{kt}$; solve

...

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.$y = e^{-x} + Ce^{-2x}$
B.$y = e^{-2x} + Ce^{-x}$
C.$y = -e^{-x} + Ce^{-2x}$
D.$y = -e^{-2x} + Ce^{-x}$

Concept Applied

Integrating factor is $e^{2x}$; multiply and integrate to get solution....

Read Full Step-by-Step Solution →

Question #48

Practice Question

Concept Applied

Separate variables: $\frac{dy}{y} = \frac{dx}{x}$, integrate: $\ln y = \ln x + C$, apply IC: $y = x$, so $y(e) = e \approx 2.72$....

Read Full Step-by-Step Solution →

Question #49

Practice Question

A.$x + y + z = 3$
B.$x^2 + y^2 + z^2 = 3$
C.$xy + yz + zx = 1$
D.$x^2 + y^2 + z^2 = 1$

Concept Applied

Step 1: Since the surface is passing through the point $(1, 1, 1)$, we can substitute these values into the equation of the surface to get

Question #50

Practice Question

A.$y = Ae^{x^2}$
B.$y = Ae^{-x^2}$
C.$y = Ae^{x^2} + B$
D.$y = Ae^{-x^2} + B$

Concept Applied

Step 1: Given the differential equation $\frac{dy}{dx} = \frac{2x}{y}$, we can rewrite it as $\frac{dy}{y} = \frac{2x}{y}dx$....

Read Full Step-by-Step Solution →