How do you differentiate y=x^(e^x) ?, №18011154, 23.08.2021 09:22

Mathematics

PLEASE HELP!! it’s multiple choice but i never get any of the answers. differentiate y=x^(e^x)

Step-by-step answer

P Answered by Master

We are given the function to differentiate:

${\quad \qquad \sf \rightarrow y=x^{e^x}}$

Do take natural log on both sides, then we will be having

${:\implies \quad \sf ln(y)=ln(x^{e^{x}})}$

${:\implies \quad \sf ln(y)=e^{x}ln(x)\quad \qquad \{\because ln(a^b)=bln(a)\}}$

Now, differentiate both sides by using so called chain rule and the product rule

${:\implies \quad \sf \dfrac{1}{y}\dfrac{dy}{dx}=e^{x}ln(x)+\dfrac{e^x}{x}}$

${:\implies \quad \boxed{\bf{\dfrac{dy}{dx}=x^{e^x}\bigg\{\dfrac{e^x}{x}+ln(x)e^{x}\bigg\}}}}$

Hence, Option B) is correct

Product rule of differentiation:

${\boxed{\bf{\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}}}}$

Where, u and v are functions of x

Mathematics

PLEASE HELP!! it’s multiple choice but i never get any of the answers. differentiate y=x^(e^x)

Step-by-step answer

P Answered by Specialist

We are given the function to differentiate:

${\quad \qquad \sf \rightarrow y=x^{e^x}}$

Do take natural log on both sides, then we will be having

${:\implies \quad \sf ln(y)=ln(x^{e^{x}})}$

${:\implies \quad \sf ln(y)=e^{x}ln(x)\quad \qquad \{\because ln(a^b)=bln(a)\}}$

Now, differentiate both sides by using so called chain rule and the product rule

${:\implies \quad \sf \dfrac{1}{y}\dfrac{dy}{dx}=e^{x}ln(x)+\dfrac{e^x}{x}}$

${:\implies \quad \boxed{\bf{\dfrac{dy}{dx}=x^{e^x}\bigg\{\dfrac{e^x}{x}+ln(x)e^{x}\bigg\}}}}$

Hence, Option B) is correct

Product rule of differentiation:

${\boxed{\bf{\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}}}}$

Where, u and v are functions of x

Mathematics

How do you differentiate y=x^(e^x) ?

Step-by-step answer

P Answered by Specialist

We are given the function to differentiate:

${\quad \qquad \sf \rightarrow y=x^{e^x}}$

Do take natural log on both sides, then we will be having

${:\implies \quad \sf ln(y)=ln(x^{e^{x}})}$

${:\implies \quad \sf ln(y)=e^{x}ln(x)\quad \qquad \{\because ln(a^b)=bln(a)\}}$

Now, differentiate both sides by using so called chain rule and the product rule

${:\implies \quad \sf \dfrac{1}{y}\dfrac{dy}{dx}=e^{x}ln(x)+\dfrac{e^x}{x}}$

${:\implies \quad \boxed{\bf{\dfrac{dy}{dx}=x^{e^x}\bigg\{\dfrac{e^x}{x}+ln(x)e^{x}\bigg\}}}}$

Hence, Option B) is correct

Product rule of differentiation:

${\boxed{\bf{\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}}}}$

Where, u and v are functions of x

Mathematics

1. (1 pt each) the equation of motion of a particle is s = t 3 − 12t 2 + 36t, t ≥ 0, where s is measured in meters and t is in seconds. (a) find the velocity at time t. (b) what is the velocity after 3 s? (c) when is the particle at rest? (d) when is the particle moving in the positive direction? (e) find the total distance traveled during the first 8 s. (f) find the acceleration at time t and after 3 s. 2. (3 pt each) given p(x) = x n − x, find the intervals over which p(x) is a decreasing function when: (a) n = 2 (b) n = 1 2 (c) n = −1 3. (a)

Step-by-step answer

P Answered by PhD

3

QUESTION 1)

The particle's equation of motion is

$s=t^3-12t^2+36t,t\ge0$

where s is measured in meters and t is in seconds.

The velocity at time, is given by the first derivative of the equation of motion of the particle.

$s'(t)=(3t^2-24t+36)ms^{-1}$

Question 1b).

To find the velocity after 3 seconds, we put t=3 into the velocity function.

$\Rightarrow s'(3)=(3(3)^2-24(3)+36)ms^{-1}$

$\Rightarrow s'(3)=(27-72+36)ms^{-1}$

$\Rightarrow s'(3)=-9ms^{-1}$

QUESTION 1C

To find the time that the particle is at rest, we equate the velocity (the first derivative formula) to zero and solve for t.

$\Rightarrow 3t^2-24t+36=0$

Divide through by 3 to get;

$\Rightarrow t^2-8t+12=0$

Factor:

$\Rightarrow (t-2)(t-6)=0$

(t-2)=0,(t-6)=0

t=2s,t=6s

The particle is at rest when t=2s and t=6s .

QUESTION 1d.

The particle is moving in a positive direction when the velocity is greater than zero.

$\Rightarrow 3t^2-24t+36\:\:0$

$\Rightarrow t^2-8t+12\:\:0$

$\Rightarrow (t-2)(t-6)\:\:0$

$\Rightarrow t\:\:6$

But $t\ge0$ .

This implies that, the particle is moving in a positive direction on the interval,

$0\le t\:\:6$

QUESTION 1e.

To find the total distance traveled after 8s, we substitute t=8 into the equation of motion of the particle.

s(8)=8^3-12(8)^2+36(8)

s(8)=512-768+288

s(8)=32m

The particle covered 32m in the first 8 seconds.

QUESTION 1f

i) The acceleration at time can be obtained by differentiating the velocity equation.

$\Rightarrow s"(t)=6t-24$

ii) To find the acceleration after 3 seconds, we substitute t=3 into the equation of acceleration.

$\Rightarrow s"(3)=6(3)-24$

$\Rightarrow s"(3)=18-24$

$\Rightarrow s"(3)=-6ms^{-2}$

After 3 seconds, the particle is decelerating at 6 meters per seconds square.

QUESTION 2a

Given:

p(x)=x^n-x

$p'(x)=nx^{n-1}-1$

When n=2 , then

$p'(x)=2x^{2-1}-1$

p'(x)=2x-1

This implies that, p(x) is decreasing when

$p'(x)\:$

$2x-1\:$

Therefore the function is decreasing on;

$x\:$

QUESTION 2b

When $n=\frac{1}{2}$

$p'(x)=\frac{1}{2\sqrt{x}}-1$

To find the interval over which the function is decreasing, we solve the inequality;

$\frac{1}{2\sqrt{x}}-1\:$

Therefore the function is decreasing on the interval;

$\Rightarrow x\:\:\frac{1}{4}$

QUESTION 3a

The given parabola has equation

y=x^2+x...(1)

Let the two tangents from the external point; (2,-3) have equation;

y+3=m(x-2)..(2)

Put equation (1) into equation (2)

This implies that;

x^2+x+3=m(x-2)

Rewrite to obtain a quadratic equation in .

x^2+(1-m)x+3+2m=0

Since this is the point of intersection of a tangent and a parabola, the discriminant of this quadratic equation must be zero.

$\Rightarrow (1-m)^2-4(3-2m)=0$

$\Rightarrow m^2-10m-11=0$

$\Rightarrow m=11\:or\:m=-1$

We substitute the values of m into equation (2) to obtain the equations of the two tangents to be;

y=11x-25 and y=-x-1

QUESTION 3b

Let the equation of the tangents from the external point (2,7) be

y-7=m(x-2)...(1)

The given parabola has equation

y=x^2+x...(2)

The discriminant of the intersection of these two equations yields;

m^2-10m+29=0

This quadratic equation has no real roots.

Hence there are no lines through the point (2,7) that are tangents to parabola.

We can see from the graph that this point is lying inside the parabola.

QUESTION 4

The given cubic function is

y=ax^3+bx^2+cx+d

$\frac{dy}{dx}=3ax^2+2bx+c$

The horizontal tangents occurs when $\frac{dy}{dx}=0$ .

$\Rightarrow 3ax^2+2bx+c=0$

This occurs at (2,0).

$\Rightarrow 12a+4b+c=0...(1)$

and (-2,6) .

$\Rightarrow 12a-4b+c=0...(2)$

These points also lie on the curve so they must satisfy the equation of the curve;

Substituting (2,0) into the original equation gives;

8a+4b+2c+d=0...(3)

Substituting (-2,6) into the original equation gives;

-8a+4b-2c+d=0...(4)

Solving the four equations simultaneously gives;

$a=\frac{3}{16},b=0,c=-\frac{9}{4},c=3$

Hence the required cubic function;

$y=\frac{3}{16}x^3-\frac{9}{4}x+3$

QUESTION 5a

Let

y=fgh

where f,g, and are differentiable.

Using the product rule;

y'=f'(gh)+f(gh)'

Use the product rule again;

y'=f'(gh)+f(g'h+gh')

y'=f'(gh)+fg'h+fgh' as required.

Mathematics

1. (1 pt each) the equation of motion of a particle is s = t 3 − 12t 2 + 36t, t ≥ 0, where s is measured in meters and t is in seconds. (a) find the velocity at time t. (b) what is the velocity after 3 s? (c) when is the particle at rest? (d) when is the particle moving in the positive direction? (e) find the total distance traveled during the first 8 s. (f) find the acceleration at time t and after 3 s. 2. (3 pt each) given p(x) = x n − x, find the intervals over which p(x) is a decreasing function when: (a) n = 2 (b) n = 1 2 (c) n = −1 3. (a)

Step-by-step answer

P Answered by PhD

3

QUESTION 1)

The particle's equation of motion is

$s=t^3-12t^2+36t,t\ge0$

where s is measured in meters and t is in seconds.

The velocity at time, is given by the first derivative of the equation of motion of the particle.

$s'(t)=(3t^2-24t+36)ms^{-1}$

Question 1b).

To find the velocity after 3 seconds, we put t=3 into the velocity function.

$\Rightarrow s'(3)=(3(3)^2-24(3)+36)ms^{-1}$

$\Rightarrow s'(3)=(27-72+36)ms^{-1}$

$\Rightarrow s'(3)=-9ms^{-1}$

QUESTION 1C

To find the time that the particle is at rest, we equate the velocity (the first derivative formula) to zero and solve for t.

$\Rightarrow 3t^2-24t+36=0$

Divide through by 3 to get;

$\Rightarrow t^2-8t+12=0$

Factor:

$\Rightarrow (t-2)(t-6)=0$

(t-2)=0,(t-6)=0

t=2s,t=6s

The particle is at rest when t=2s and t=6s .

QUESTION 1d.

The particle is moving in a positive direction when the velocity is greater than zero.

$\Rightarrow 3t^2-24t+36\:\:0$

$\Rightarrow t^2-8t+12\:\:0$

$\Rightarrow (t-2)(t-6)\:\:0$

$\Rightarrow t\:\:6$

But $t\ge0$ .

This implies that, the particle is moving in a positive direction on the interval,

$0\le t\:\:6$

QUESTION 1e.

To find the total distance traveled after 8s, we substitute t=8 into the equation of motion of the particle.

s(8)=8^3-12(8)^2+36(8)

s(8)=512-768+288

s(8)=32m

The particle covered 32m in the first 8 seconds.

QUESTION 1f

i) The acceleration at time can be obtained by differentiating the velocity equation.

$\Rightarrow s"(t)=6t-24$

ii) To find the acceleration after 3 seconds, we substitute t=3 into the equation of acceleration.

$\Rightarrow s"(3)=6(3)-24$

$\Rightarrow s"(3)=18-24$

$\Rightarrow s"(3)=-6ms^{-2}$

After 3 seconds, the particle is decelerating at 6 meters per seconds square.

QUESTION 2a

Given:

p(x)=x^n-x

$p'(x)=nx^{n-1}-1$

When n=2 , then

$p'(x)=2x^{2-1}-1$

p'(x)=2x-1

This implies that, p(x) is decreasing when

$p'(x)\:$

$2x-1\:$

Therefore the function is decreasing on;

$x\:$

QUESTION 2b

When $n=\frac{1}{2}$

$p'(x)=\frac{1}{2\sqrt{x}}-1$

To find the interval over which the function is decreasing, we solve the inequality;

$\frac{1}{2\sqrt{x}}-1\:$

Therefore the function is decreasing on the interval;

$\Rightarrow x\:\:\frac{1}{4}$

QUESTION 3a

The given parabola has equation

y=x^2+x...(1)

Let the two tangents from the external point; (2,-3) have equation;

y+3=m(x-2)..(2)

Put equation (1) into equation (2)

This implies that;

x^2+x+3=m(x-2)

Rewrite to obtain a quadratic equation in .

x^2+(1-m)x+3+2m=0

Since this is the point of intersection of a tangent and a parabola, the discriminant of this quadratic equation must be zero.

$\Rightarrow (1-m)^2-4(3-2m)=0$

$\Rightarrow m^2-10m-11=0$

$\Rightarrow m=11\:or\:m=-1$

We substitute the values of m into equation (2) to obtain the equations of the two tangents to be;

y=11x-25 and y=-x-1

QUESTION 3b

Let the equation of the tangents from the external point (2,7) be

y-7=m(x-2)...(1)

The given parabola has equation

y=x^2+x...(2)

The discriminant of the intersection of these two equations yields;

m^2-10m+29=0

This quadratic equation has no real roots.

Hence there are no lines through the point (2,7) that are tangents to parabola.

We can see from the graph that this point is lying inside the parabola.

QUESTION 4

The given cubic function is

y=ax^3+bx^2+cx+d

$\frac{dy}{dx}=3ax^2+2bx+c$

The horizontal tangents occurs when $\frac{dy}{dx}=0$ .

$\Rightarrow 3ax^2+2bx+c=0$

This occurs at (2,0).

$\Rightarrow 12a+4b+c=0...(1)$

and (-2,6) .

$\Rightarrow 12a-4b+c=0...(2)$

These points also lie on the curve so they must satisfy the equation of the curve;

Substituting (2,0) into the original equation gives;

8a+4b+2c+d=0...(3)

Substituting (-2,6) into the original equation gives;

-8a+4b-2c+d=0...(4)

Solving the four equations simultaneously gives;

$a=\frac{3}{16},b=0,c=-\frac{9}{4},c=3$

Hence the required cubic function;

$y=\frac{3}{16}x^3-\frac{9}{4}x+3$

QUESTION 5a

Let

y=fgh

where f,g, and are differentiable.

Using the product rule;

y'=f'(gh)+f(gh)'

Use the product rule again;

y'=f'(gh)+f(g'h+gh')

y'=f'(gh)+fg'h+fgh' as required.

Advanced Placement (AP)

Which of the following is the solution to the differential equation dy/dx=e^(y+x) with initial condition y(0) = -ln4 A) y= -x-ln4 B) y=x-ln4 C) y = -ln(-e^x+5) D) y = -ln(e^x+3) E) y = ln(e^x+3)

Step-by-step answer

P Answered by Master

C) y = -ln(-eˣ + 5)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Algebra I

Function NotationExponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$ Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Log PropertiesNatural log ln(x) and eˣ

Calculus

Antiderivatives - Integrals

Integration Constant C

U-Substitution

Slope Fields

Solving DifferentialsSeparation of Variables

Explanation:

Step 1: Define

$\displaystyle \frac{dy}{dx} = e^{y + x} \\y(0) = -ln4$

Step 2: Rewrite

Separation of Variables. Get differential equation to a form where we can integrate both sides.

[Differential Equation] Rewrite [Exponential Rule - Multiplying]: $\displaystyle \frac{dy}{dx} = e^y \cdot e^x$ [Diff Eq] Isolate x terms together [Multiplication Property of Equality]: $\displaystyle dy = e^y \cdot e^x dx$ [Diff Eq] Isolate y terms together [Division Property of Equality]: $\displaystyle \frac{dy}{e^y} = e^x dx$ [Diff Eq] Rewrite: $\displaystyle \frac{1}{e^y} dy = e^x dx$ [Diff Eq] Rewrite y [Exponential Rule - Rewrite]: $\displaystyle e^{-y} dy = e^x dx$

Step 3: Integrate Pt. 1

[Diff Eq] Integrate both sides [Equality Property]: $\displaystyle \int {e^{-y}} \, dy = \int {e^x} \, dx$

Step 4: Identify Variables for U-Substitution

Set variables for u-sub for y.

u = -y

du = -dy

Step 5: Integrate Pt. 2

[Integrals] Rewrite: $\displaystyle -\int {-e^{-y}} \, dy = \int {e^x} \, dx$ [Integrals] U-Substitution: $\displaystyle -\int {e^u} \, du = \int {e^x} \, dx$ [Integrals] eˣ integration: $\displaystyle -e^u = e^x + C$ [Integral Expression] Back-substitution: $\displaystyle -e^{-y} = e^x + C$

Step 6: Solve Differential Equation Pt. 1

[Int Exp] Divide -1 on both sides [Division Property of Equality]: $\displaystyle e^{-y} = -e^x - C$ [Int Exp] Natural log both sides (isolate y term) [Equality Property]: $\displaystyle -y = ln(-e^x - C)$ [Int Exp] Divide -1 on both sides [Division Property of Equality]: $\displaystyle y = -ln(-e^x - C)$

This is our differential function.

Step 7: Solve Differential Equation Pt. 2

[Diff Function] Substitute in given point: $\displaystyle -ln4 = -ln(-e^0 - C)$ [Diff Function] Evaluate exponent: $\displaystyle -ln4 = -ln(-1 - C)$ [Diff Function] Divide -1 on both sides [Division Property of Equality]: $\displaystyle ln4 = ln(-1 - C)$ [Diff Function] e both sides [Equality Property]: $\displaystyle 4 = -1 - C$ [Diff Function] Add 1 on both sides [Addition Property of Equality]: $\displaystyle 5 = -C$ [Diff Function] Divide -1 on both sides [Division Property of Equality]: $\displaystyle -5 = C$ [Diff Function] Rewrite: $\displaystyle C = -5$ [Diff Function] Substitute in Integration Constant C: $\displaystyle y = -ln(-e^x - -5)$ [Diff Function] Simplify: $\displaystyle y = -ln(-e^x + 5)$

Topic: Calculus

Unit: Slope Fields

Book: College Calculus 10e

Advanced Placement (AP)

Which of the following is the solution to the differential equation dy/dx=e^(y+x) with initial condition y(0) = -ln4 A) y= -x-ln4 B) y=x-ln4 C) y = -ln(-e^x+5) D) y = -ln(e^x+3) E) y = ln(e^x+3)

Step-by-step answer

P Answered by Master

C) y = -ln(-eˣ + 5)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Algebra I

Function NotationExponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$ Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Log PropertiesNatural log ln(x) and eˣ

Calculus

Antiderivatives - Integrals

Integration Constant C

U-Substitution

Slope Fields

Solving DifferentialsSeparation of Variables

Explanation:

Step 1: Define

$\displaystyle \frac{dy}{dx} = e^{y + x} \\y(0) = -ln4$

Step 2: Rewrite

Separation of Variables. Get differential equation to a form where we can integrate both sides.

[Differential Equation] Rewrite [Exponential Rule - Multiplying]: $\displaystyle \frac{dy}{dx} = e^y \cdot e^x$ [Diff Eq] Isolate x terms together [Multiplication Property of Equality]: $\displaystyle dy = e^y \cdot e^x dx$ [Diff Eq] Isolate y terms together [Division Property of Equality]: $\displaystyle \frac{dy}{e^y} = e^x dx$ [Diff Eq] Rewrite: $\displaystyle \frac{1}{e^y} dy = e^x dx$ [Diff Eq] Rewrite y [Exponential Rule - Rewrite]: $\displaystyle e^{-y} dy = e^x dx$

Step 3: Integrate Pt. 1

[Diff Eq] Integrate both sides [Equality Property]: $\displaystyle \int {e^{-y}} \, dy = \int {e^x} \, dx$

Step 4: Identify Variables for U-Substitution

Set variables for u-sub for y.

u = -y

du = -dy

Step 5: Integrate Pt. 2

[Integrals] Rewrite: $\displaystyle -\int {-e^{-y}} \, dy = \int {e^x} \, dx$ [Integrals] U-Substitution: $\displaystyle -\int {e^u} \, du = \int {e^x} \, dx$ [Integrals] eˣ integration: $\displaystyle -e^u = e^x + C$ [Integral Expression] Back-substitution: $\displaystyle -e^{-y} = e^x + C$

Step 6: Solve Differential Equation Pt. 1

[Int Exp] Divide -1 on both sides [Division Property of Equality]: $\displaystyle e^{-y} = -e^x - C$ [Int Exp] Natural log both sides (isolate y term) [Equality Property]: $\displaystyle -y = ln(-e^x - C)$ [Int Exp] Divide -1 on both sides [Division Property of Equality]: $\displaystyle y = -ln(-e^x - C)$

This is our differential function.

Step 7: Solve Differential Equation Pt. 2

[Diff Function] Substitute in given point: $\displaystyle -ln4 = -ln(-e^0 - C)$ [Diff Function] Evaluate exponent: $\displaystyle -ln4 = -ln(-1 - C)$ [Diff Function] Divide -1 on both sides [Division Property of Equality]: $\displaystyle ln4 = ln(-1 - C)$ [Diff Function] e both sides [Equality Property]: $\displaystyle 4 = -1 - C$ [Diff Function] Add 1 on both sides [Addition Property of Equality]: $\displaystyle 5 = -C$ [Diff Function] Divide -1 on both sides [Division Property of Equality]: $\displaystyle -5 = C$ [Diff Function] Rewrite: $\displaystyle C = -5$ [Diff Function] Substitute in Integration Constant C: $\displaystyle y = -ln(-e^x - -5)$ [Diff Function] Simplify: $\displaystyle y = -ln(-e^x + 5)$

Topic: Calculus

Unit: Slope Fields

Book: College Calculus 10e

Mathematics

Consider the following differential equation. x^2y + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) + C)/x is a solution of the differential equation. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen. (Do this on paper. Your instructor may ask you to turn in this work.) (c) Find a solution of the differential equation that satisfies the initial condition y(1) = 3. (Enter the argument of the logarithmic

Step-by-step answer

P Answered by Master

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

x^2y' +xy = 3

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

Mathematics

Consider the following differential equation. x^2y + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) + C)/x is a solution of the differential equation. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Illustrate part (a) by graphing several members of the family of solutions on a common screen. (Do this on paper. Your instructor may ask you to turn in this work.) (c) Find a solution of the differential equation that satisfies the initial condition y(1) = 3. (Enter the argument of the logarithmic

Step-by-step answer

P Answered by Specialist

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

x^2y' +xy = 3

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

Mathematics

Consider the differential equation: y − 2y + 17y = 0; e^x cos 4x, ex sin 4x, (−[infinity], [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent since W(e^x cos 4x, e^x sin 4x) = ≠ 0 for −[infinity] x [infinity]. Form the general solution. y =

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P Answered by Specialist

The question is:

Consider the differential equation:

$y''-2y' + 17y = 0 \\ e^xcos 4x, e^x sin 4x\\ $on the interval$ (-\infty, \infty).$

(1) Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent since $W\left(e^x cos 4x, e^x sin 4x \right) \neq 0 \\ $on$ -\infty< x < \infty.$

(2) Form the general solution.

(1) To verify if the given functions form a fundamental set of solutions to the differential equation, we find the Wronskian of the two functions.

The Wronskian of functions y_1 $and$ y_2 is given as

$W(y_1, y_2) = \left|\begin{array}{cc}y_1&y_2\\y_1'&y_2'\end{array}\right|\\\\y_1= e^x cos 4x \\y_2 = e^x sin 4x \\y_1' = -4e^x sin 4x \\y_2' = 4e^x cos 4x \\$

$W\left(e^x cos 4x, e^x sin 4x \right) = \left|\begin{array}{cc}e^x cos 4x &e^x sin 4x \\ \\ -4e^x sin 4x&4e^x cos 4x \end{array}\right|$

$= 4e^{2x} cos^2 4x + e^{2x} sin^2 4x \\ \\ = 4e^{2x}\left( cos^2 4x + sin^2 4x\right)\\ \\$but $cos^2 4x + sin^2 4x = 1\\ \\W\left(y_1, y_2 \right) = 4e^{2x} \neq 0$

(2) The general solution may be expressed as a linear combination

y = C_1e^x cos 4x + C_2e^x sin 4x

Where C_1 $ and $ C_2 are arbitrary constants.

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