Swift Tuition's Swift Guide to Tutorial 7

Solution:

We need to verify that when we substitute $y = \frac{\ln x + C}{x}$ into the equation, we get a true statement.

Step 1: Find $y'$ using the quotient rule.

For $y = \frac{\ln x + C}{x}$:

$$y' = \frac{x \cdot \frac{d}{dx}(\ln x + C) - (\ln x + C) \cdot \frac{d}{dx}(x)}{x^2}$$ $$y' = \frac{x \cdot \frac{1}{x} - (\ln x + C) \cdot 1}{x^2}$$ $$y' = \frac{1 - \ln x - C}{x^2}$$

Step 2: Calculate $x^2y'$.

$$x^2y' = x^2 \cdot \frac{1 - \ln x - C}{x^2} = 1 - \ln x - C$$

Step 3: Calculate $xy$.

$$xy = x \cdot \frac{\ln x + C}{x} = \ln x + C$$

Step 4: Verify $x^2y' + xy = 1$.

$$x^2y' + xy = (1 - \ln x - C) + (\ln x + C) = 1$$

The equation is satisfied.

Solution:

Step 1: Find the first derivative $y'$.

Using the product rule on each term:

For $ax^2\cos x$:

$$\frac{d}{dx}(ax^2\cos x) = a[2x\cos x - x^2\sin x]$$

For $bx\sin x$:

$$\frac{d}{dx}(bx\sin x) = b[\sin x + x\cos x]$$

Therefore:

$$y' = a(2x\cos x - x^2\sin x) + b(\sin x + x\cos x)$$

Step 2: Find the second derivative $y''$.

For the first term $a(2x\cos x - x^2\sin x)$:

$$\frac{d}{dx}[a(2x\cos x - x^2\sin x)] = a[(2\cos x - 2x\sin x) - (2x\sin x + x^2\cos x)]$$ $$= a[2\cos x - 4x\sin x - x^2\cos x]$$

For the second term $b(\sin x + x\cos x)$:

$$\frac{d}{dx}[b(\sin x + x\cos x)] = b[\cos x + (\cos x - x\sin x)]$$ $$= b[2\cos x - x\sin x]$$

Therefore:

$$y'' = a(2\cos x - 4x\sin x - x^2\cos x) + b(2\cos x - x\sin x)$$

Step 3: Calculate $y'' + y$.

Original: $y = ax^2\cos x + bx\sin x$

$$y'' + y = a(2\cos x - 4x\sin x - x^2\cos x) + b(2\cos x - x\sin x) + ax^2\cos x + bx\sin x$$

Collecting terms:

• Coefficient of $x^2\cos x$: $-a + a = 0$
• Coefficient of $\cos x$: $2a + 2b$
• Coefficient of $x\sin x$: $-4a - b + b = -4a$
• Coefficient of $\sin x$: $0$

So: $y'' + y = (2a + 2b)\cos x - 4ax\sin x$

Step 4: Match with $x\sin x$.

We need: $(2a + 2b)\cos x - 4ax\sin x = x\sin x$

This gives us:

• Coefficient of $\cos x$: $2a + 2b = 0$
• Coefficient of $x\sin x$: $-4a = 1$

From the second equation: $a = -\frac{1}{4}$

From the first equation: $2(-\frac{1}{4}) + 2b = 0$, so $b = \frac{1}{4}$

Answer: $a = -\frac{1}{4}$, $b = \frac{1}{4}$

Step 1: Convert to standard linear form.

Divide by $x^2$:

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

This is now in the form $y' + P(x)y = Q(x)$ where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{x^2}$.

Step 2: Find the integrating factor.

$$I(x) = e^{\int P(x)dx} = e^{\int \frac{1}{x}dx} = e^{\ln|x|} = |x|$$

We take $I(x) = x$ (assuming $x > 0$).

Step 3: Multiply the equation by $I(x) = x$.

$$x \cdot y' + x \cdot \frac{1}{x}y = x \cdot \frac{1}{x^2}$$ $$xy' + y = \frac{1}{x}$$

Step 4: Recognize that the left side is $(xy)'$.

By the product rule: $(xy)' = xy' + y$

So our equation becomes:

$$(xy)' = \frac{1}{x}$$

Step 5: Integrate both sides.

$$xy = \int \frac{1}{x}dx = \ln|x| + C$$

Step 6: Solve for $y$.

$$y = \frac{\ln|x| + C}{x}$$

Step 1: Recognize this as separable.

We can write: $\frac{dy}{dx} = y^2 + 1$

Step 2: Separate variables.

$$\frac{dy}{y^2 + 1} = dx$$

Step 3: Integrate both sides.

$$\int \frac{dy}{y^2 + 1} = \int dx$$

The left integral is a standard form:

$$\arctan(y) = x + C$$

Step 4: Solve for $y$.

$$y = \tan(x + C)$$

Step 1: Factor the right side.

$$\frac{dy}{dt} = t^2y - y + t^2 - 1 = y(t^2 - 1) + 1(t^2 - 1) = (t^2 - 1)(y + 1)$$

Step 2: Separate variables.

$$\frac{dy}{y + 1} = (t^2 - 1)dt$$

Step 3: Integrate both sides.

$$\int \frac{dy}{y + 1} = \int (t^2 - 1)dt$$ $$\ln|y + 1| = \frac{t^3}{3} - t + C$$

Step 4: Solve for $y$.

$$|y + 1| = e^{\frac{t^3}{3} - t + C}$$

Step 1: Factor the right side.

$$\frac{dy}{dx} = 3x^2(2 - y)$$

Step 2: Separate variables.

$$\frac{dy}{2 - y} = 3x^2dx$$

Step 3: Integrate both sides.

For the left side, use substitution. Let $u = 2 - y$, then $du = -dy$:

$$\int \frac{dy}{2 - y} = -\int \frac{du}{u} = -\ln|u| = -\ln|2 - y|$$

For the right side:

$$\int 3x^2dx = x^3 + C_1$$

So: $-\ln|2 - y| = x^3 + C_1$

Step 4: Solve for $y$.

$$\ln|2 - y| = -x^3 - C_1$$ $$|2 - y| = e^{-x^3 - C_1} = e^{-C_1} \cdot e^{-x^3}$$

Let $K = e^{-C_1}$ (a positive constant):

$$|2 - y| = K e^{-x^3}$$ $$2 - y = \pm K e^{-x^3}$$

Since $\pm K$ can be any non-zero constant, we write it as $C$:

$$2 - y = C e^{-x^3}$$ $$y = 2 - C e^{-x^3}$$

Alternatively (changing the sign of $C$):

$$y = 2 + C e^{-x^3}$$

Step 1: Convert to standard form.

Divide by $x^4$:

$$y' + \frac{4}{x}y = \frac{\sin^3 x}{x^4}$$

Step 2: Find the integrating factor.

$$I(x) = e^{\int \frac{4}{x}dx} = e^{4\ln|x|} = e^{\ln|x|^4} = |x|^4 = x^4$$

Step 3: Multiply by the integrating factor.

$$x^4y' + 4x^3y = \sin^3 x$$

The left side is $(x^4y)'$, so:

$$(x^4y)' = \sin^3 x$$

Step 4: Integrate $\sin^3 x$.

We use the identity: $\sin^3 x = \sin x(1 - \cos^2 x) = \sin x - \sin x\cos^2 x$

First integral: $\int \sin x\, dx = -\cos x$

Second integral: For $\int \sin x\cos^2 x\, dx$, let $u = \cos x$, so $du = -\sin x\, dx$:

$$\int \sin x\cos^2 x\, dx = -\int u^2\, du = -\frac{u^3}{3} = -\frac{\cos^3 x}{3}$$

Therefore:

$$\int \sin^3 x\, dx = -\cos x - \left(-\frac{\cos^3 x}{3}\right) = -\cos x + \frac{\cos^3 x}{3}$$

Step 5: Complete the solution.

$$x^4y = -\cos x + \frac{\cos^3 x}{3} + C$$ $$y = -x^{-4}\cos x + \frac{1}{3}x^{-4}\cos^3 x + x^{-4}C$$

Step 1: Convert to standard linear form.

$$y' - y = -x$$

Here, $P(x) = -1$ and $Q(x) = -x$.

Step 2: Find the integrating factor.

$$I(x) = e^{\int -1\, dx} = e^{-x}$$

Step 3: Multiply by the integrating factor.

$$e^{-x}y' - e^{-x}y = -xe^{-x}$$

The left side is $(e^{-x}y)'$:

$$(e^{-x}y)' = -xe^{-x}$$

Step 4: Integrate the right side using integration by parts.

For $\int -xe^{-x}dx$:

Let $u = -x$ and $dv = e^{-x}dx$

Then $du = -dx$ and $v = -e^{-x}$

$$\int -xe^{-x}dx = (-x)(-e^{-x}) - \int (-e^{-x})(-dx)$$ $$= xe^{-x} - \int e^{-x}dx$$ $$= xe^{-x} + e^{-x} + C$$ $$= (x + 1)e^{-x} + C$$

Step 5: Solve for $y$.

$$e^{-x}y = (x + 1)e^{-x} + C$$ $$y = x + 1 + Ce^x$$

Step 1: Convert to standard form.

Divide by $x^2$:

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

Step 2: Find the integrating factor.

$$I(x) = e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = e^{\ln|x|^2} = x^2$$

Step 3: Multiply by the integrating factor.

$$x^2y' + 2xy = \ln x$$

The left side is $(x^2y)'$:

$$(x^2y)' = \ln x$$

Step 4: Integrate using integration by parts.

For $\int \ln x\, dx$:

Let $u = \ln x$ and $dv = dx$

Then $du = \frac{1}{x}dx$ and $v = x$

$$\int \ln x\, dx = x\ln x - \int x \cdot \frac{1}{x}dx = x\ln x - \int dx = x\ln x - x + C$$

Step 5: Apply the integration result.

$$x^2y = x\ln x - x + C$$

Step 6: Apply the initial condition $y(1) = 2$.

$$1^2 \cdot 2 = 1 \cdot \ln 1 - 1 + C$$ $$2 = 0 - 1 + C$$ $$C = 3$$

Step 7: Write the final solution.

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

Step 1: Rearrange as a separable equation.

$$x\ln x\, dx = y(1 + y^2)dy$$

Step 2: Integrate both sides.

Left side (using integration by parts):

$$\int x\ln x\, dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C_1$$

Right side:

$$\int y(1 + y^2)dy = \int (y + y^3)dy = \frac{y^2}{2} + \frac{y^4}{4} + C_2$$

Step 3: Combine and apply initial condition.

$$\frac{x^2}{2}\ln x - \frac{x^2}{4} = \frac{y^2}{2} + \frac{y^4}{4} + C$$

At $x = 1$, $y = 1$:

$$\frac{1}{2}\ln 1 - \frac{1}{4} = \frac{1}{2} + \frac{1}{4} + C$$ $$0 - \frac{1}{4} = \frac{3}{4} + C$$ $$C = -1$$

The solution is given implicitly by:

$$\frac{x^2}{2}\ln x - \frac{x^2}{4} = \frac{y^2}{2} + \frac{y^4}{4} - 1$$

Step 1: Separate variables.

$$\frac{dy}{y} = x\sin x\, dx$$

Step 2: Integrate both sides.

Left side:

$$\int \frac{dy}{y} = \ln|y|$$

Right side (using integration by parts):

Let $u = x$ and $dv = \sin x\, dx$

Then $du = dx$ and $v = -\cos x$

$$\int x\sin x\, dx = -x\cos x - \int (-\cos x)dx = -x\cos x + \sin x + C$$

Step 3: Combine the results.

$$\ln|y| = -x\cos x + \sin x + C$$

Step 4: Apply the initial condition $y(0) = -1$.

$$\ln|-1| = -0 \cdot \cos 0 + \sin 0 + C$$ $$\ln 1 = 0 + 0 + C$$ $$C = 0$$

Step 5: Solve for $y$.

$$\ln|y| = -x\cos x + \sin x$$

Since $y(0) = -1 < 0$, we have $y < 0$ throughout (solutions don't cross $y = 0$), so:

$$\ln(-y) = -x\cos x + \sin x$$ $$-y = e^{-x\cos x + \sin x}$$ $$y = -e^{-x\cos x + \sin x}$$

Step 1: Convert to standard form.

Divide by $x$:

$$y' = \frac{y}{x} + x\sin x$$ $$y' - \frac{1}{x}y = x\sin x$$

Step 2: Find the integrating factor.

$$I(x) = e^{\int -\frac{1}{x}dx} = e^{-\ln|x|} = e^{\ln|x|^{-1}} = \frac{1}{|x|} = \frac{1}{x}$$

Step 3: Multiply by the integrating factor.

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

The left side is $\left(\frac{y}{x}\right)'$:

$$\left(\frac{y}{x}\right)' = \sin x$$

Step 4: Integrate.

$$\frac{y}{x} = -\cos x + C$$

Step 5: Apply the initial condition $y(\pi) = 0$.

$$\frac{0}{\pi} = -\cos \pi + C$$ $$0 = -(-1) + C$$ $$C = -1$$

Step 6: Write the final solution.

$$\frac{y}{x} = -\cos x - 1$$ $$y = -x(\cos x + 1)$$

Step 1: Set up the mass balance equation.

The rate of change of CO₂ mass = (CO₂ flowing in) - (CO₂ flowing out)

If $c(t)$ is the concentration at time $t$:

• CO₂ mass in room = $V \cdot c(t) = 180c(t)$
• CO₂ flowing in per minute = $Q \cdot c_{in} = 2 \times 0.0005 = 0.001$ m³/min
• CO₂ flowing out per minute = $Q \cdot c(t) = 2c(t)$ m³/min

$$180\frac{dc}{dt} = 0.001 - 2c(t)$$

Step 2: Simplify the equation.

$$\frac{dc}{dt} = \frac{0.001 - 2c}{180} = \frac{1}{90}(0.0005 - c)$$

Step 3: Solve the separable equation.

$$\frac{dc}{0.0005 - c} = -\frac{1}{90}dt$$

Integrating:

$$-\ln|0.0005 - c| = -\frac{t}{90} + K$$

Step 4: Apply the initial condition $c(0) = 0.0015$.

$$-\ln|0.0005 - 0.0015| = 0 + K$$ $$-\ln(0.001) = K$$ $$K = \ln(1000)$$

Step 5: Solve for $c(t)$.

$$-\ln|0.0005 - c| = -\frac{t}{90} + \ln(1000)$$ $$\ln|0.0005 - c| = \frac{t}{90} - \ln(1000)$$

Since $c(0) = 0.0015 > 0.0005$, we have $c > 0.0005$ initially, so:

$$c - 0.0005 = 0.001e^{-t/90}$$ $$c(t) = 0.0005 + 0.001e^{-t/90}$$

Step 6: Convert to percentage.

$$\text{CO}_2(t) = 100c(t) = 0.05 + 0.10e^{-t/90} \text{ percent}$$

As $t \to \infty$, the concentration approaches 0.05%.

Forces acting on the object:

• Downward: Weight = $mg$
• Upward: Air resistance = $kv$ (proportional to velocity)

By Newton's Second Law ($F = ma$):

$$m\frac{dv}{dt} = mg - kv$$

Rearranging:

$$\frac{dv}{dt} + \frac{k}{m}v = g$$

This is a linear first-order equation with $P = \frac{k}{m}$ and $Q = g$.

Integrating factor: $I = e^{\int \frac{k}{m}dt} = e^{kt/m}$

Multiplying by $I$:

$$(e^{kt/m}v)' = ge^{kt/m}$$

Integrating:

$$e^{kt/m}v = g \cdot \frac{m}{k}e^{kt/m} + C$$

If $v(0) = 0$ (starts from rest):

$$C = -\frac{mg}{k}$$

Therefore:

$$v(t) = \frac{mg}{k}(1 - e^{-kt/m})$$

As $t \to \infty$, $e^{-kt/m} \to 0$, so:

$$v_{\infty} = \frac{mg}{k}$$

This is when the drag force equals the weight.

$$s(t) = \int_0^t v(\tau)d\tau = \int_0^t \frac{mg}{k}(1 - e^{-k\tau/m})d\tau$$ $$s(t) = \frac{mg}{k}\left[t + \frac{m}{k}e^{-kt/m}\right]_0^t$$ $$s(t) = \frac{mg}{k}\left(t + \frac{m}{k}e^{-kt/m} - \frac{m}{k}\right)$$

Step 1: Differentiate the general solution.

$$y' = 3x^2 - 3Cx^{-4}$$

Step 2: Eliminate the arbitrary constant $C$.

From the original equation: $Cx^{-3} = y - x^3$

So: $C = x^3(y - x^3) = x^3y - x^6$

Substituting into the derivative:

$$y' = 3x^2 - 3(x^3y - x^6)x^{-4}$$ $$y' = 3x^2 - 3x^{-1}y + 3x^2$$ $$y' = 6x^2 - \frac{3y}{x}$$

Multiplying by $x$:

$$xy' = 6x^3 - 3y$$

Rearranging:

$$xy' + 3y = 6x^3$$

Step 1: Find $f'(x)$ for any $x$.

By definition of derivative:

$$f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$

Using the functional equation $f(x+h) = f(x)f(h)$:

$$f'(x) = \lim_{h \to 0}\frac{f(x)f(h) - f(x)}{h} = f(x) \cdot \lim_{h \to 0}\frac{f(h) - 1}{h}$$

Note that $\lim_{h \to 0}\frac{f(h) - 1}{h} = \lim_{h \to 0}\frac{f(0+h) - f(0)}{h} = f'(0) = 1$

Therefore: $f'(x) = f(x)$

Step 2: Solve the differential equation.

We have the initial value problem:

• $f'(x) = f(x)$
• $f(0) = 1$

This is a separable equation:

$$\frac{df}{f} = dx$$

Integrating:

$$\ln|f| = x + C$$

Since $f(0) = 1$:

$$\ln 1 = 0 + C$$ $$C = 0$$

Therefore:

$$\ln|f| = x$$ $$f(x) = e^x$$

(We take the positive solution since $f(0) = 1 > 0$)