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Q7 Find the general solution of: \[ 2 t \cdot \frac{d y}{d t}-y=t+1 \text {, wi...
May 15, 2024
Solution by Steps
step 1
Given the differential equation 2t⋅dydt−y=t+12t \cdot \frac{dy}{dt} - y = t + 1 with the initial condition y(2)=4y(2) = 4, we first rewrite it in standard form
step 2
Rewrite the equation as dydt−y2t=t+12t\frac{dy}{dt} - \frac{y}{2t} = \frac{t + 1}{2t}
step 3
Identify the integrating factor μ(t)=e∫−12tdt=e−12ln⁡t=t−12 \mu(t) = e^{\int -\frac{1}{2t} dt} = e^{-\frac{1}{2} \ln t} = t^{-\frac{1}{2}}
step 4
Multiply both sides of the differential equation by the integrating factor t−12 t^{-\frac{1}{2}} : t−12dydt−t−12⋅y2t=t−12⋅t+12t t^{-\frac{1}{2}} \frac{dy}{dt} - t^{-\frac{1}{2}} \cdot \frac{y}{2t} = t^{-\frac{1}{2}} \cdot \frac{t + 1}{2t} which simplifies to ddt(t−12y)=t12+t−122 \frac{d}{dt} \left( t^{-\frac{1}{2}} y \right) = \frac{t^{\frac{1}{2}} + t^{-\frac{1}{2}}}{2}
step 5
Integrate both sides with respect to t t : t−12y=∫t12+t−122dt t^{-\frac{1}{2}} y = \int \frac{t^{\frac{1}{2}} + t^{-\frac{1}{2}}}{2} dt t−12y=∫t122dt+∫t−122dt t^{-\frac{1}{2}} y = \int \frac{t^{\frac{1}{2}}}{2} dt + \int \frac{t^{-\frac{1}{2}}}{2} dt t−12y=12⋅23t32+12⋅2t12+C t^{-\frac{1}{2}} y = \frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} + \frac{1}{2} \cdot 2 t^{\frac{1}{2}} + C t−12y=13t32+t12+C t^{-\frac{1}{2}} y = \frac{1}{3} t^{\frac{3}{2}} + t^{\frac{1}{2}} + C
step 6
Multiply both sides by t12 t^{\frac{1}{2}} to solve for y y : y=t⋅13t32+t⋅t12+Ct12 y = t \cdot \frac{1}{3} t^{\frac{3}{2}} + t \cdot t^{\frac{1}{2}} + C t^{\frac{1}{2}} y=13t2+t+Ct12 y = \frac{1}{3} t^2 + t + C t^{\frac{1}{2}}
step 7
Use the initial condition y(2)=4 y(2) = 4 to find C C : 4=13⋅4+2+C⋅2 4 = \frac{1}{3} \cdot 4 + 2 + C \cdot \sqrt{2} 4=43+2+C⋅2 4 = \frac{4}{3} + 2 + C \cdot \sqrt{2} 4=103+C⋅2 4 = \frac{10}{3} + C \cdot \sqrt{2} 4−103=C⋅2 4 - \frac{10}{3} = C \cdot \sqrt{2} 123−103=C⋅2 \frac{12}{3} - \frac{10}{3} = C \cdot \sqrt{2} 23=C⋅2 \frac{2}{3} = C \cdot \sqrt{2} C=232=226=23 C = \frac{2}{3 \sqrt{2}} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}
step 8
Substitute C C back into the general solution: y=13t2+t+23t12 y = \frac{1}{3} t^2 + t + \frac{\sqrt{2}}{3} t^{\frac{1}{2}}
Answer
y=13t2+t+23t12 y = \frac{1}{3} t^2 + t + \frac{\sqrt{2}}{3} t^{\frac{1}{2}}
Key Concept
First-order linear ordinary differential equation
Explanation
The solution involves finding an integrating factor, rewriting the equation, integrating, and applying the initial condition to find the constant.
Solve each questions
Solution by Steps
step 1
Given the differential equation
dθdt=−k⋅(θ−10)\frac{d \theta}{d t} = -k \cdot (\theta - 10)
and the initial condition
θ(0)=70∘C\theta(0) = 70^\circ \mathrm{C}
, we need to find the constant $k$.
step 2
Separate the variables and integrate:
∫1θ−10 dθ=−k∫ dt\int \frac{1}{\theta - 10} \, d\theta = -k \int \, dt
.
step 3
Integrate both sides:
ln⁡∣θ−10∣=−kt+C\ln|\theta - 10| = -kt + C
.
step 4
Use the initial condition
θ(0)=70\theta(0) = 70
to find $C$:
ln⁡∣70−10∣=C  ⟹  C=ln⁡60\ln|70 - 10| = C \implies C = \ln 60
.
step 5
Use the condition
θ(10)=40\theta(10) = 40
to find $k$:
ln⁡∣40−10∣=−10k+ln⁡60  ⟹  ln⁡30=−10k+ln⁡60\ln|40 - 10| = -10k + \ln 60 \implies \ln 30 = -10k + \ln 60
.
step 6
Solve for kk:
−10k=ln⁡3060=ln⁡12=−ln⁡2  ⟹  k=ln⁡210-10k = \ln \frac{30}{60} = \ln \frac{1}{2} = -\ln 2 \implies k = \frac{\ln 2}{10}
.
# Part (b) Find the body's temperature after further 15 minutes
step 1
Use the value of kk found in part (a) and the general solution:
θ(t)=10+Ce−kt\theta(t) = 10 + Ce^{-kt}
.
step 2
Substitute the initial condition to find CC:
70=10+60e0  ⟹  C=6070 = 10 + 60e^{0} \implies C = 60
.
step 3
Find the temperature after 25 minutes (10 + 15):
θ(25)=10+60e−ln⁡210⋅25=10+60e−2510ln⁡2=10+60e−2.5ln⁡2=10+60⋅2−2.5\theta(25) = 10 + 60e^{-\frac{\ln 2}{10} \cdot 25} = 10 + 60e^{-\frac{25}{10} \ln 2} = 10 + 60e^{-2.5 \ln 2} = 10 + 60 \cdot 2^{-2.5}
.
step 4
Simplify the expression:
θ(25)=10+60⋅122.5=10+60⋅132=10+60⋅142=10+60⋅14⋅1.414≈10+60⋅0.1768≈10+10.608≈20.608∘C\theta(25) = 10 + 60 \cdot \frac{1}{2^{2.5}} = 10 + 60 \cdot \frac{1}{\sqrt{32}} = 10 + 60 \cdot \frac{1}{4\sqrt{2}} = 10 + 60 \cdot \frac{1}{4 \cdot 1.414} \approx 10 + 60 \cdot 0.1768 \approx 10 + 10.608 \approx 20.608^\circ \mathrm{C}
.
Answer
k=ln⁡210k = \frac{\ln 2}{10}, θ(25)≈20.608∘C\theta(25) \approx 20.608^\circ \mathrm{C}
Question 9
step 1
The moment of inertia of a triangle about its base is given by
I=13bh3I = \frac{1}{3}bh^3
.
step 2
For a triangle with height
h=2 mh = 2 \, \mathrm{m}
and base
b=1 mb = 1 \, \mathrm{m}
, substitute these values into the formula:
I=13⋅1⋅(2)3=13⋅8=83 m4I = \frac{1}{3} \cdot 1 \cdot (2)^3 = \frac{1}{3} \cdot 8 = \frac{8}{3} \, \mathrm{m}^4
.
Answer
83 m4\frac{8}{3} \, \mathrm{m}^4
Question 10
step 1
The cumulative distribution function (CDF) is found by integrating the probability density function (PDF):
F(t)=∫0tf(x) dxF(t) = \int_0^t f(x) \, dx
.
step 2
Given the PDF
f(t)=3t2f(t) = 3t^2
for
0 < t < 1
, integrate to find the CDF:
F(t)=∫0t3x2 dx=[x3]0t=t3F(t) = \int_0^t 3x^2 \, dx = [x^3]_0^t = t^3
.
step 3
Verify by long-hand integration:
∫0t3x2 dx=3∫0tx2 dx=3[x33]0t=t3\int_0^t 3x^2 \, dx = 3 \int_0^t x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_0^t = t^3
.
Answer
F(t)=t3F(t) = t^3 for 0 < t < 1
Key Concept
Differential Equations, Moments of Inertia, Cumulative Distribution Function
Explanation
The solutions involve solving a first-order linear differential equation, calculating the moment of inertia for a triangular cross-section, and integrating a probability density function to find the cumulative distribution function.
Solve these three questions
Solution by Steps
step 1
Given the bending moment equation for Beam A: M=3⋅x2+2⋅cos⁡xM = 3 \cdot \frac{x}{2} + 2 \cdot \cos x
step 2
To find the load distribution, we need to find the second derivative of MM with respect to xx
step 3
First derivative: dMdx=32−2sin⁡x\frac{dM}{dx} = \frac{3}{2} - 2 \sin x
step 4
Second derivative: d2Mdx2=−2cos⁡x\frac{d^2M}{dx^2} = -2 \cos x
step 5
The load distribution for Beam A is w(x)=−d2Mdx2=2cos⁡xw(x) = -\frac{d^2M}{dx^2} = 2 \cos x
step 6
To find the point of maximum bending moment, set dMdx=0\frac{dM}{dx} = 0: 32−2sin⁡x=0\frac{3}{2} - 2 \sin x = 0
step 7
Solving for xx: sin⁡x=34\sin x = \frac{3}{4}
# Beam B
step 1
Given the bending moment equation for Beam B: M=4ln⁡x+7sin⁡x−3exM = 4 \ln x + 7 \sin x - 3 e^x
step 2
First derivative: dMdx=4x+7cos⁡x−3ex\frac{dM}{dx} = \frac{4}{x} + 7 \cos x - 3 e^x
step 3
Second derivative: d2Mdx2=−4x2−7sin⁡x−3ex\frac{d^2M}{dx^2} = -\frac{4}{x^2} - 7 \sin x - 3 e^x
step 4
The load distribution for Beam B is w(x)=−d2Mdx2=4x2+7sin⁡x+3exw(x) = -\frac{d^2M}{dx^2} = \frac{4}{x^2} + 7 \sin x + 3 e^x
Answer
Load distribution for Beam A: w(x)=2cos⁡xw(x) = 2 \cos x
Load distribution for Beam B: w(x)=4x2+7sin⁡x+3exw(x) = \frac{4}{x^2} + 7 \sin x + 3 e^x
Point of maximum bending moment for Beam A: sin⁡x=34\sin x = \frac{3}{4}
Key Concept
Bending Moment and Load Distribution
Explanation
The load distribution is found by taking the second derivative of the bending moment equation. The point of maximum bending moment is found by setting the first derivative to zero.
Question 2: Logarithmic Differentiation
step 1
Given the function y=(sin⁡x−x)x+1y = (\sin x - x)^{x+1}
step 2
Take the natural logarithm of both sides: ln⁡y=(x+1)ln⁡(sin⁡x−x)\ln y = (x+1) \ln (\sin x - x)
step 3
Differentiate both sides with respect to xx: 1ydydx=(x+1)ddxln⁡(sin⁡x−x)+ln⁡(sin⁡x−x)ddx(x+1)\frac{1}{y} \frac{dy}{dx} = (x+1) \frac{d}{dx} \ln (\sin x - x) + \ln (\sin x - x) \frac{d}{dx} (x+1)
step 4
Simplify the derivatives: ddxln⁡(sin⁡x−x)=cos⁡x−1sin⁡x−x\frac{d}{dx} \ln (\sin x - x) = \frac{\cos x - 1}{\sin x - x} and ddx(x+1)=1\frac{d}{dx} (x+1) = 1
step 5
Combine the results: 1ydydx=(x+1)cos⁡x−1sin⁡x−x+ln⁡(sin⁡x−x)\frac{1}{y} \frac{dy}{dx} = (x+1) \frac{\cos x - 1}{\sin x - x} + \ln (\sin x - x)
step 6
Multiply both sides by yy: dydx=y[(x+1)cos⁡x−1sin⁡x−x+ln⁡(sin⁡x−x)]\frac{dy}{dx} = y \left[ (x+1) \frac{\cos x - 1}{\sin x - x} + \ln (\sin x - x) \right]
step 7
Substitute back y=(sin⁡x−x)x+1y = (\sin x - x)^{x+1}: dydx=(sin⁡x−x)x+1[(x+1)cos⁡x−1sin⁡x−x+ln⁡(sin⁡x−x)]\frac{dy}{dx} = (\sin x - x)^{x+1} \left[ (x+1) \frac{\cos x - 1}{\sin x - x} + \ln (\sin x - x) \right]
Answer
dydx=(sin⁡x−x)x+1[(x+1)cos⁡x−1sin⁡x−x+ln⁡(sin⁡x−x)]\frac{dy}{dx} = (\sin x - x)^{x+1} \left[ (x+1) \frac{\cos x - 1}{\sin x - x} + \ln (\sin x - x) \right]
Key Concept
Logarithmic Differentiation
Explanation
Logarithmic differentiation is used to simplify the differentiation of functions where the variable is both in the base and the exponent.
Question 3: Partial Derivative
step 1
Given the function r=y2+x2−6x−4y+8r = y^2 + x^2 - 6x - 4y + 8
step 2
To find the partial derivative of rr with respect to xx, treat yy as a constant
step 3
Differentiate rr with respect to xx: ∂r∂x=2x−6\frac{\partial r}{\partial x} = 2x - 6
Answer
∂r∂x=2x−6\frac{\partial r}{\partial x} = 2x - 6
Key Concept
Partial Derivative
Explanation
The partial derivative of a function with respect to one variable is found by differentiating with respect to that variable while treating all other variables as constants.
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