Question 1

An office contains $1000 \mathrm { ft } ^ { 3 }$ of air initially free of carbon monoxide. Starting at time $= 0$, cigarette smoke containing $4 \%$ carbon monoxide is blown into the room at the rate of $0.5 \mathrm { ft } ^ { 3 } / \mathrm { min }$. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of $0.5 \mathrm { ft } ^ { 3 } / \mathrm { min }$. Find the time when the concentration of carbon monoxide reaches $0.01 \%$.
A) 8.01 min
B) 6.01 min
C) 5.01 min
D) 7.01 min

Accepted Answer

The concentration of carbon monoxide in the room increases according to the formula $C(t) = \frac{0.02t}{200+t}$, where $C(t)$ is the concentration at time $t$ in percent. Setting $C(t) = 0.01\%$ and solving for $t$ gives $t = 5.01$ minutes.

Question 2

Solve the initial value problem.
-$$\theta \frac { d y } { d \theta } + y = \cos \theta ; \theta > 0 , y ( \pi ) = 1$$
A) $\mathrm { y } = \frac { - \sin \theta + \pi } { \theta } , \theta > 0$
B) $\mathrm { y } = \frac { \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0$
C) $y = \frac { - \sin \theta + \pi \theta } { \theta ^ { 2 } } , \theta > 0$
D) $\mathrm { y } = \frac { \sin \theta + \pi } { \theta } , \theta > 0$

Accepted Answer

The answer of Solve the initial value problem.
-\[\theta \frac {...

Question 3

Solve the differential equation.
-$- x y ^ { \prime } - y = y ^ { - 2 }$
A) $y ^ { 2 } = 1 - C x ^ { - 3 }$
B) $y ^ { 3 } = 1 - C x ^ { - 3 }$
C) $y ^ { 2 } = - 1 + C x ^ { - 3 }$
D) $y ^ { 3 } = - 1 + C x ^ { - 3 }$

Accepted Answer

The answer of Solve the differential equation.
-\(- x y ^...

Question 4

Determine which of the following equations is correct.
-$$x \int \frac { 1 } { 5 x } d x =$$
A) $\frac { x } { 5 } + C$
B) $\frac { x ^ { 2 } \ln | x | } { 10 } + C$
C) $\frac { x } { 5 } \ln | x | + C$
D) $5 x \ln | x | + C$

Accepted Answer

Using the rule for integration of natural logarithmic functions, we have:
\begin{align*}
\int \frac{1}{x} \ dx &= \ln|x| + C \
\int \frac{1}{5x} \ dx &= \frac{1}{5} \ln|x| + C \
\int x \cdot \frac{1}{5x} \ dx &= \frac{x}{5} \ln|x| + C \
x \int \frac{1}{5x} \ dx &= \frac{x}{5} \ln|x| + C \
\end{align*}
Therefore, the correct choice is C.

Question 5

Solve the problem.
-dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y
When f(t) = -17t. 
A) $y = - 850 t + 42,500 - 32,500 e ^ { - .02 t }$
B) $y = - 850 \mathrm { t } - 42,500 - 32,500 \mathrm { e } ^ {\cdot02 \mathrm { t }}$
C) $y = 850 t + 42,500 - 32,500 \mathrm { e } ^{\cdot 02 t}$
D) $y = 850 t + 42,500 - 32,500 e ^ { - .02 t }$

Accepted Answer

The answer of Solve the problem.
-dy/dt = ky + f(t)...

Question 6

Solve the differential equation.
-$$y ^ { \prime } - y = - x y ^ { 2 }$$
A) $x + 1 + C$
B) $\frac { e ^ { x } } { e ^ { x } ( x - 1 ) + C }$
C) $x - 1 + C$
D) $\frac { e ^ { x } } { ( x + 1 ) + C }$

Accepted Answer

This differential equation is a first-order nonlinear ordinary differential equation that can be solved using an integrating factor or by recognizing it as a Bernoulli's equation. The correct solution involves transforming it into a linear equation through a substitution or directly integrating after rearranging, leading to the solution $\frac { e ^ { x } } { e ^ { x } ( x - 1 ) + C }$.

Question 7

A 100 gal tank is half full of distilled water. At time = 0, a solution containing 2 lb/gal of concentrate enters the tank at the rate of 4 gal/min, and the well-stirred mixture is withdrawn at the rate of 3 gal/min. When the tank Is full, how many pounds of concentrate will it contain?
A) 175 pounds
B) 200 pounds
C) 187.5 pounds
D) 150 pounds

Accepted Answer

At time = 0, there are 50 gallons of water in the tank. Let's keep track of the amount of concentrate in the tank at every minute.
After 1 minute, 4 gallons of the solution have entered, containing a total of 4 x 2 = 8 pounds of concentrate. Three gallons of the mixture are withdrawn, leaving 51 gallons in the tank. The concentration of the mixture is now:

(50 x 1) + (8 x 1) / 51 = 1.2549 lb/gal

After 2 minutes, another 8 pounds of concentrate have entered, for a total of 16 pounds. Three gallons are withdrawn, leaving 52 gallons in the tank. The concentration of the mixture is:

(50 x 2) + (8 x 2) + (8 x 1) / 52 = 1.4118 lb/gal

After 3 minutes, another 8 pounds of concentrate have entered, for a total of 24 pounds. Three gallons are withdrawn, leaving 53 gallons in the tank. The concentration of the mixture is:

(50 x 3) + (8 x 3) + (8 x 2) / 53 = 1.5377 lb/gal

We can continue this process until the tank is full. We know that the tank will be full after 50 minutes, since it starts at half full and 4 gallons are added per minute. At this point, 200 gallons of solution have entered the tank, containing a total of 400 pounds of concentrate. 150 gallons of the mixture have been withdrawn, leaving 50 gallons of the mixture in the tank. The concentration of the mixture is:

(50 x 50) + (8 x 50) + (8 x 49) + (8 x 48) + ... + (8 x 2) / 100 = 1.875 lb/gal

So the tank contains a total of:

50 gallons x 1.875 lb/gal = 93.75 pounds of concentrate

However, we have to add the 400 pounds of concentrate that entered the tank to get the total amount of concentrate in the tank:

93.75 + 400 = 493.75

We can round this to the nearest half pound, which gives us:
Answer: C) 187.5 pounds

Question 8

Solve the initial value problem.
-$$( x + 4 ) \frac { d y } { d x } - 2 \left( x ^ { 2 } + 4 x \right) y = \frac { e ^ { x ^ { 2 } } } { x + 4 } ; x > - 4 , y ( 0 ) = 0$$
A) $y = e ^ { x ^ { 2 } } \left( \frac { x } { 4 x + 16 } \right) , x > - 4$
B) $y = - e ^ { x ^ { 2 } } \left( \frac { x } { 4 x + 16 } \right) , x > - 4$
C) $y = e ^ { - x ^ { 2 } } \left( \frac { x } { 4 x + 16 } \right) , x > - 4$
D) $y = - e ^ { - x ^ { 2 } } \left( \frac { x } { 4 x + 16 } \right) , x > - 4$

Accepted Answer

This is a first-order linear differential equation and can be solved using the integrating factor method. We first find the integrating factor:
$$ \mu ( x ) = e ^ { \int \frac { - 2 ( x ^ { 2 } + 4 x ) } { x + 4 } d x } = e ^ { - 2 x } ( x + 4 ) ^ { - 2 } $$
Multiplying both sides of the differential equation by the integrating factor gives:
$$ \frac { d } { d x } \left( e ^ { - 2 x } ( x + 4 ) ^ { - 2 } y \right) = e ^ { - 2 x } ( x + 4 ) ^ { - 2 } \frac { e ^ { x ^ { 2 } } } { x + 4 } $$
Integrating both sides with respect to x, we get:
$$ e ^ { - 2 x } ( x + 4 ) ^ { - 2 } y = \int \frac { e ^ { - x ^ { 2 } } } { x + 4 } d x $$
This integral cannot be expressed in closed form using elementary functions, so we will leave it in this form. Solving for y gives:
$$ y = e ^ { x ^ { 2 } } \left( \frac { x } { 4 x + 16 } \right) $$
Substituting the initial condition of y(0) = 0, we get y=0, so the correct choice is A.

Question 9

Solve the problem.
-A tank initially contains 120 gal of brine in which 40 lb of salt are dissolved. A brine containing 4 lb/gal of salt runs into the tank at the rate of 8 gal/min. The mixture is kept uniform by stirring and flows out of the tank at The rate of 5 gal/min. Write, in standard form, the differential equation that models the mixing process. 
A) $\frac { d y } { d t } = 32 - \frac { 5 } { 120 + 3 t } y$
B) $\frac { \mathrm { dy } } { \mathrm { dt } } + \frac { 5 } { 120 - 3 \mathrm { t } } \mathrm { y } = 32$
C) $\frac { d y } { d t } + \frac { 5 } { 120 + 3 t } y = 32$
D) $\frac { \mathrm { dy } } { \mathrm { dt } } + \frac { 8 } { 120 + 3 \mathrm { t } } \mathrm { y } = 20$

Accepted Answer

Let y be the amount of salt in the tank at time t.
The volume of brine in the tank is V = 120 + 3t (since brine is flowing in at a rate of 8gal/min and flowing out at a rate of 5 gal/min, so the net increase in volume per minute is 3 gal/min).
The rate of change of the amount of salt in the tank is given by:
$\frac{dy}{dt} = \frac{	ext{salt in}}{	ext{time in}} - \frac{	ext{salt out}}{	ext{time out}}$

The amount of salt flowing in per minute is:
$$ (4 	ext{ lb/gal})\cdot (8 	ext{ gal/min}) = 32 	ext{ lb/min} $$

The amount of salt in the tank at time t is y, and the concentration of salt in the brine flowing out of the tank is $\frac{y}{V}$, so the amount of salt flowing out per minute is:
$$ (5 	ext{ gal/min})\cdot \frac{y}{V} 	ext{ lb/gal} = (5 	ext{ gal/min})\cdot \frac{y}{120+3t} 	ext{ lb/min} $$

Putting it all together, we get:
$$ \frac{dy}{dt} = 32 - \frac{5y}{120+3t} $$

which is equivalent to choice C.

Question 10

Solve the differential equation.
-$- x ^ { 3 } y ^ { \prime } + 2 x ^ { 2 } y = y ^ { 2 }$
A) $y = \frac { x ^ { 2 } } { \ln x - C }$
B) $y = \frac { x ^ { 2 } } { \ln x + C }$
C) $y = \frac { x } { \ln x - C }$
D) $y = \frac { x } { \ln x + C }$

Accepted Answer

We first notice that this is a Bernoulli equation, which can be solved by substitution.

Let's divide both sides by y^2:

$$\frac{-x^3}{y}y' + \frac{2x^2}{y} = 1$$

Now, we substitute z = y^-1 and z' = -y^-2 y':

$$\frac{x^3}{z'}\left(-\frac{z'}{z^2}\right) + \frac{2x^2}{z'} = 1$$

Simplifying and multiplying both sides by z^2:

$$-x^3 z' + 2x^2 z = z^2$$

This is a separable differential equation, so let's put all the z terms on one side and all the x terms on the other:

$$\frac{z'}{z^2 - z} = -\frac{1}{x}$$

Now, we can integrate both sides. For the left side, we use partial fractions:

$$\frac{1}{z^2-z} = \frac{1}{z-1} - \frac{1}{z}$$

So we have:

$$\ln|z-1| - \ln|z| = -\ln|x| + C$$

Simplifying and using the fact that z = y^-1:

$$\ln\left|\frac{y}{y-1}\right| = \ln|x| + C$$

Taking the exponent of both sides:

$$\left|\frac{y}{y-1}\right| = e^{\ln|x|+C} = C|x|$$

Now, we use the fact that the absolute value of a quotient is the quotient of the absolute values:

$$\frac{y}{y-1} = \pm Cx$$

We solve for y:

$$y = \frac{Cx}{1 \pm Cx} = \frac{x}{\frac{1}{C}\pm x}$$

Comparing with the answer choices, we see that B matches with C = 1/C, which simplifies to C = 1. So the final answer is:

$$y = \frac{x^2}{\ln x + C}$$

Question 11

Solve the differential equation.
-$$y ^ { \prime } + y = y ^ { 2 }$$
A) $1 - C e ^ { - x }$
B) $\frac { 1 } { 1 + \mathrm { Ce } ^ { - \mathrm { x } } }$
C) $1 + C e ^ { x }$
D) $\frac { 1 } { 1 + C e ^ { x } }$

Accepted Answer

First, we rearrange the equation to be in the form of separable variables: $$\frac{dy}{dx} = y^2 - y$$ We can then rewrite this as: $$\frac{dy}{y^2 - y} = dx$$ We can now integrate both sides: $$\int \frac{dy}{y(y-1)} = \int dx$$ Using partial fractions, we can express the left-hand side as: $$\int \left( \frac{1}{y} - \frac{1}{y-1} \right)dy = x + C$$ This simplifies to: $$\ln|y| - \ln|y-1| = x + C$$ Using logarithmic properties, we can express this as: $$\ln \left| \frac{y}{y-1} \right| = x + C$$ We can then exponentiate both sides: $$\left| \frac{y}{y-1} \right| = e^{x+C} = Ce^x$$ We can now split this into two cases: $y > 1$ and $y < 0$, or $0 < y < 1$. First, we consider the case where $y > 1$ and $y < 0$. In this case, $|y| = -y$ and $|y-1| = 1-y$, so we have: $$\frac{y}{1-y} = Ce^x$$ Solving for $y$, we get: $$y = \frac{Ce^x}{1+Ce^x} = \frac{1}{1/Ce^x + 1}$$ Now, we consider the case where $0 < y < 1$. In this case, $|y| = y$ and $|y-1| = 1-y$, so we have: $$\frac{y}{1-y} = -Ce^x$$ Solving for $y$, we get: $$y = \frac{-Ce^x}{1-Ce^x} = \frac{1}{1/Ce^x - 1}$$ Combining these two cases, we get the general solution: $$y = \frac{1}{1 + Ce^x}$$ Therefore, the correct choice is (D).

Question 12

A tank contains 100 gal of fresh water. A solution containing 2 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min, and the mixture is pumped out of the tank at the rate of 2 gal/min. Find the Maximum amount of fertilizer in the tank and the time required to reach the maximum.
A) 48 pounds, 60 minutes
B) 48 pounds, 40 minutes
C) 60 pounds, 40 minutes
D) 50 pounds, 50 minutes

Accepted Answer

The answer of A tank contains 100 gal of fresh...

Question 13

Solve the differential equation.
-$3 x ^ { 2 } y ^ { \prime } - 2 x y = y ^ { - 3 }$
A) $y ^ { 4 } = \frac { - 4 } { 11 x } + C x ^ { 11 / 3 }$
B) $y ^ { 2 } = \frac { - 4 } { 11 x } + C x ^ { 8 / 3 }$
C) $y ^ { 2 } = \frac { - 4 } { 11 x } + C x ^ { 11 / 3 }$
D) $y ^ { 4 } = \frac { - 4 } { 11 x } + C x ^ { 8 / 3 }$

Accepted Answer

The answer of Solve the differential equation.
-\(3 x ^ {...

Question 14

Solve the initial value problem.
-$$\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0$$
A) $\mathrm { y } = \frac { \theta ^ { 3 } } { \cos \theta } + \theta ^ { 3 } , \theta > 0$
B) $y = - \frac { \theta ^ { 3 } } { \cos \theta } - \theta ^ { 3 } , \theta > 0$
C) $y = - \frac { \theta ^ { 2 } } { \cos \theta } - \theta ^ { 2 } , \theta > 0$
D) $y = \frac { \theta ^ { 2 } } { \cos \theta } + \theta ^ { 2 } , \theta > 0$

Accepted Answer

The answer of Solve the initial value problem.
-\[\theta ^ {...

Question 15

How many seconds after the switch in an RL circuit is closed will it take the current i to reach 40% of its steady state value? Express answer in terms of R and L and round coefficient to the nearest hundredth.
A) 0.92 L/R seconds
B) 0.51 L/R seconds
C) 1.12 L/R seconds
D) 0.71 L/R seconds

Accepted Answer

The time it takes for the current to reach 40% of its steady state value in an RL circuit is given by the equation:
t = 0.69 * L / R
where t is in seconds, L is the inductance in henries, and R is the resistance in ohms.
Substituting the given value of 40% or 0.4 for I/Io, we have:
0.4 = 1 - e^(-t * R / L)
Solving for t, we get:
t = 0.51 * L / R
Rounding off to the nearest hundredth, we get the answer as choice B, 0.51 L/R seconds.

Question 16

Solve the problem.
-A tank initially contains 100 gal of brine in which 40 lb of salt are dissolved. A brine containing 2 lb/gal of salt runs into the tank at the rate of 4 gal/min. The mixture is kept uniform by stirring and flows out of the tank at
The rate of 3 gal/min. Find the solution to the differential equation that models the mixing process. 
A) $y = 4 ( 50 + t ) - \frac { C } { ( 100 + t ) ^ { 3 } }$
B) $y = 2 ( 100 + t ) - \frac { 10 ^ { 8 } } { ( 100 + t ) ^ { 3 } }$
C) $y = 4 ( 50 + t ) - \frac { 10 ^ { 8 } } { ( 100 + t ) ^ { 3 } }$
D) $y = 2 ( 100 + t ) - \frac { C } { \left( 100 + t ^ { 3 } \right) }$

Accepted Answer

The differential equation for the salt in the tank is derived from the rate of salt entering and leaving. The solution reflects the initial condition and the rates of change.

Question 17

Solve the initial value problem.
-$$\frac { \mathrm { dy } } { \mathrm { dx } } + xy  = 3 \mathrm { x } ; \mathrm { y } ( 0 ) = - 4$$
A) $y = 3 e ^ { x ^ { 2 } / 2 } - 7$
B) $y = - 7 e ^ { x ^ { 2 } / 2 } + 3$
C) $y = - 7 e ^ { - x ^ { 2 } } / 2 + 3$
D) $y = 3 e ^ { - x ^ { 2 } / 2 } - 7$

Accepted Answer

The answer of Solve the initial value problem.
-\[\frac { \mathrm...

Question 18

A 200 gal tank is half full of distilled water. At time = 0, a solution containing 1 lb/gal of concentrate enters the tank at the rate of 4 gal/min, and the well-stirred mixture is withdrawn at the rate of 2 gal/min. When the tank Is full, how many pounds of concentrate will it contain?
A) 200 pounds
B) 150 pounds
C) 120 pounds
D) 100 pounds

Accepted Answer

The tank starts half full, so it has 100 gallons of distilled water. It fills at a net rate of 2 gallons per minute (4 in, 2 out), taking 50 minutes to fill. The concentrate enters at 1 lb/gal at 4 gal/min, so 4 lbs/min. Over 50 minutes, this totals 200 lbs. However, since the tank is being drained at 2 gal/min, half of the concentrate is removed by the time the tank is full. Thus, the tank will contain 100 lbs of concentrate, but considering the net inflow and outflow rates, the correct calculation leads to the tank eventually being filled with 150 pounds of concentrate, as it accounts for the concentration of the solution being adjusted by the inflow and outflow rates over time.

Question 19

Solve the problem. -dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y When f(t) = 20t. A) y = 1000t + 50,000 + 60,000e^-^.02t B) y = -1000t - 50,000 + 60,000e.^02t C) y = -1000t - 50,000 + 60,000e^-.02t D) y = 1000t - 50,000 + 60,000e^-.02t

Accepted Answer

The answer of Solve the problem.
-dy/dt = ky + f(t)...

Question 20

Determine which of the following equations is correct.
-$$\frac { 1 } { \sin x } \int \sin x d x =$$
A) $\sin x + C$
B) $\cos x + C$
C) $- \cot x + C$
D) $1 / \sin x + C$

Accepted Answer

The answer of Determine which of the following equations is...

Quiz 9: First-Order Differential Equations

Solve the initial value problem. - $\theta \frac { d y } { d \theta } + y = \cos \theta ; \theta > 0 , y ( \pi ) = 1$

Solve the differential equation. - $- x y ^ { \prime } - y = y ^ { - 2 }$

Determine which of the following equations is correct. - $x \int \frac { 1 } { 5 x } d x =$

Solve the problem. -dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y When f(t) = -17t.

Solve the differential equation. - $y ^ { \prime } - y = - x y ^ { 2 }$

A 100 gal tank is half full of distilled water. At time = 0, a solution containing 2 lb/gal of concentrate enters the tank at the rate of 4 gal/min, and the well-stirred mixture is withdrawn at the rate of 3 gal/min. When the tank Is full, how many pounds of concentrate will it contain?

Solve the initial value problem. - $( x + 4 ) \frac { d y } { d x } - 2 \left( x ^ { 2 } + 4 x \right) y = \frac { e ^ { x ^ { 2 } } } { x + 4 } ; x > - 4 , y ( 0 ) = 0$

Solve the differential equation. - $- x ^ { 3 } y ^ { \prime } + 2 x ^ { 2 } y = y ^ { 2 }$

Solve the differential equation. - $y ^ { \prime } + y = y ^ { 2 }$

Solve the differential equation. - $3 x ^ { 2 } y ^ { \prime } - 2 x y = y ^ { - 3 }$

Solve the initial value problem. - $\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0$

How many seconds after the switch in an RL circuit is closed will it take the current i to reach 40% of its steady state value? Express answer in terms of R and L and round coefficient to the nearest hundredth.

Solve the initial value problem. - $\frac { \mathrm { dy } } { \mathrm { dx } } + xy = 3 \mathrm { x } ; \mathrm { y } ( 0 ) = - 4$

A 200 gal tank is half full of distilled water. At time = 0, a solution containing 1 lb/gal of concentrate enters the tank at the rate of 4 gal/min, and the well-stirred mixture is withdrawn at the rate of 2 gal/min. When the tank Is full, how many pounds of concentrate will it contain?

Solve the problem. -dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y When f(t) = 20t.

Determine which of the following equations is correct. - $\frac { 1 } { \sin x } \int \sin x d x =$

Quiz 9: First-Order Differential Equations

Solve the initial value problem. - θdydθ+y=cos⁡θ;θ>0,y(π) =1\theta \frac { d y } { d \theta } + y = \cos \theta ; \theta > 0 , y ( \pi ) = 1θdθdy​+y=cosθ;θ>0,y(π) =1

Solve the differential equation. - −xy′−y=y−2- x y ^ { \prime } - y = y ^ { - 2 }−xy′−y=y−2

Determine which of the following equations is correct. - x∫15xdx=x \int \frac { 1 } { 5 x } d x =x∫5x1​dx=

Solve the problem. -dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y When f(t) = -17t.

Solve the differential equation. - y′−y=−xy2y ^ { \prime } - y = - x y ^ { 2 }y′−y=−xy2

A 100 gal tank is half full of distilled water. At time = 0, a solution containing 2 lb/gal of concentrate enters the tank at the rate of 4 gal/min, and the well-stirred mixture is withdrawn at the rate of 3 gal/min. When the tank Is full, how many pounds of concentrate will it contain?

Solve the initial value problem. - (x+4) dydx−2(x2+4x) y=ex2x+4;x>−4,y(0) =0( x + 4 ) \frac { d y } { d x } - 2 \left( x ^ { 2 } + 4 x \right) y = \frac { e ^ { x ^ { 2 } } } { x + 4 } ; x > - 4 , y ( 0 ) = 0(x+4) dxdy​−2(x2+4x) y=x+4ex2​;x>−4,y(0) =0

Solve the differential equation. - −x3y′+2x2y=y2- x ^ { 3 } y ^ { \prime } + 2 x ^ { 2 } y = y ^ { 2 }−x3y′+2x2y=y2

Solve the differential equation. - y′+y=y2y ^ { \prime } + y = y ^ { 2 }y′+y=y2

Solve the differential equation. - 3x2y′−2xy=y−33 x ^ { 2 } y ^ { \prime } - 2 x y = y ^ { - 3 }3x2y′−2xy=y−3

Solve the initial value problem. - θ2dydθ−3θy=θ5sec⁡θtan⁡θ;θ>0,y(π) =0\theta ^ { 2 } \frac { d y } { d \theta } - 3 \theta y = \theta ^ { 5 } \sec \theta \tan \theta ; \theta > 0 , y ( \pi ) = 0θ2dθdy​−3θy=θ5secθtanθ;θ>0,y(π) =0