Differential Equations and Modeling

Given: $x \dfrac{dy}{dx} + 2y = x^2$

Concept: The *order* of a differential equation is the order of the highest-order derivative present. The *degree* is the power of that highest-order derivative after the equation is made free of radicals and fractions in derivatives.

Working:
- The highest-order derivative present is $\dfrac{dy}{dx}$ (first derivative).
- It appears with power 1.

Answer: Order = 1, Degree = 1

Given: $\dfrac{dy}{dx} + e^y = 0$

Concept: Order = order of highest derivative; Degree = power of highest derivative (must be a polynomial in derivatives).

Working:
- The highest-order derivative is $\dfrac{dy}{dx}$ (first derivative), appearing with power 1.
- The term $e^y$ involves $y$ (not a derivative), so it does not affect the degree.

Answer: Order = 1, Degree = 1

Given: $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0$

Working:
- The highest-order derivative is $\dfrac{d^2y}{dx^2}$ (second derivative).
- It appears with power 1.

Answer: Order = 2, Degree = 1

Given: $\left(\dfrac{dy}{dx}\right)^4 + 3y\left(\dfrac{d^3y}{dx^3}\right) = 0$

Working:
- Derivatives present: $\dfrac{dy}{dx}$ (order 1) and $\dfrac{d^3y}{dx^3}$ (order 3).
- Highest-order derivative: $\dfrac{d^3y}{dx^3}$ — this is order 3.
- The highest-order derivative $\dfrac{d^3y}{dx^3}$ appears with power 1.

Answer: Order = 3, Degree = 1

*(Note: The answer key states order 2, degree 1, but based on the equation as written the highest derivative is third order. If the second term were $\dfrac{d^2y}{dx^2}$, the answer would be order 2, degree 1. Students should follow the equation as printed.)*

Given: $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$

As stated in the problem: $y' = \dfrac{dy}{dx}$, $y'' = \dfrac{d^2y}{dx^2}$, $y''' = \dfrac{d^3y}{dx^3}$.

Working:
- Derivatives present: $y'$ (order 1), $y''$ (order 2), $y'''$ (order 3).
- Highest-order derivative: $y'''$ — order = 3.
- The highest-order derivative $y'''$ appears as $(y''')^2$, so its power = 2.
- The equation is already a polynomial in derivatives.

Answer: Order = 3, Degree = 2

Given: $y = ae^{-x}$

Step 1 – Differentiate:
$$\frac{dy}{dx} = -ae^{-x}$$

Step 2 – Substitute into the LHS of the DE:
$$\frac{dy}{dx} + y = -ae^{-x} + ae^{-x} = 0 = \text{RHS}$$

Conclusion: Since LHS = RHS, $y = ae^{-x}$ is a solution of $\dfrac{dy}{dx} + y = 0$. ✓

Given: $y = \sqrt{1+x^2} = (1+x^2)^{1/2}$

Step 1 – Differentiate:
$$\frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$$

Step 2 – Compute RHS:
$$\frac{xy}{1+x^2} = \frac{x \cdot \sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$$

Step 3 – Compare:
$$\text{LHS} = \frac{x}{\sqrt{1+x^2}} = \text{RHS}$$

Conclusion: $y = \sqrt{1+x^2}$ is a solution. ✓

Given: $xy = \log y + c$

Step 1 – Differentiate implicitly with respect to $x$:
$$y + x\frac{dy}{dx} = \frac{1}{y}\frac{dy}{dx}$$

Step 2 – Collect $\dfrac{dy}{dx}$ terms:
$$y = \frac{dy}{dx}\left(\frac{1}{y} - x\right) = \frac{dy}{dx}\cdot\frac{1-xy}{y}$$

Step 3 – Solve for $\dfrac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{y \cdot y}{1-xy} = \frac{y^2}{1-xy}$$

Conclusion: This equals the RHS of the given DE. Hence $xy = \log y + c$ is a solution. ✓

Given: $ax^2 + by^2 = 1$ … (i)

Step 1 – First differentiation (w.r.t. $x$):
$$2ax + 2by\,y_1 = 0 \implies ax + by\,y_1 = 0 \quad \dots (ii)$$

Step 2 – Second differentiation:
$$a + b(y_1^2 + y\,y_2) = 0 \quad \dots (iii)$$

Step 3 – Eliminate $a$ and $b$.

From (ii): $a = -\dfrac{by\,y_1}{x}$

Substitute into (iii):
$$-\frac{by\,y_1}{x} + b(y_1^2 + y\,y_2) = 0$$

Divide by $b$ (assuming $b \neq 0$):
$$-\frac{y\,y_1}{x} + y_1^2 + y\,y_2 = 0$$

Multiply throughout by $x$:
$$-y\,y_1 + x\,y_1^2 + xy\,y_2 = 0$$
$$\Rightarrow x(y\,y_2 + y_1^2) = y\,y_1$$

Conclusion: This is exactly the given DE. Hence $ax^2 + by^2 = 1$ is a solution. ✓

Given: $y = (a+bx)e^{2x}$

Step 1 – First derivative:
$$y_1 = b\,e^{2x} + 2(a+bx)e^{2x} = e^{2x}[b + 2(a+bx)] = e^{2x}(2a+b+2bx)$$

Step 2 – Second derivative:
$$y_2 = 2e^{2x}(2a+b+2bx) + e^{2x}(2b) = e^{2x}(4a+2b+4bx+2b) = e^{2x}(4a+4b+4bx)$$

Step 3 – Substitute into $y_2 - 4y_1 + 4y$:
$$y_2 - 4y_1 + 4y = e^{2x}(4a+4b+4bx) - 4e^{2x}(2a+b+2bx) + 4e^{2x}(a+bx)$$
$$= e^{2x}\left[(4a+4b+4bx) - (8a+4b+8bx) + (4a+4bx)\right]$$
$$= e^{2x}\left[4a+4b+4bx - 8a-4b-8bx+4a+4bx\right]$$
$$= e^{2x}\left[(4a-8a+4a)+(4b-4b)+(4bx-8bx+4bx)\right]$$
$$= e^{2x}[0+0+0] = 0$$

Conclusion: $y = (a+bx)e^{2x}$ is a solution of $y_2 - 4y_1 + 4y = 0$. ✓

Given: $x^2 = 2y^2 \log y$ … (i)

Step 1 – Differentiate implicitly w.r.t. $x$:
$$2x = 2\left[2y\log y \cdot \frac{dy}{dx} + y^2 \cdot \frac{1}{y}\cdot\frac{dy}{dx}\right]$$
$$2x = 2\frac{dy}{dx}\left[2y\log y + y\right]$$
$$x = \frac{dy}{dx}\cdot y(2\log y + 1)$$

$$\Rightarrow \frac{dy}{dx} = \frac{x}{y(2\log y+1)} \quad \dots (ii)$$

Step 2 – From (i): $2\log y = \dfrac{x^2}{y^2}$, so $2\log y + 1 = \dfrac{x^2+y^2}{y^2}$.

Step 3 – Substitute into (ii):
$$\frac{dy}{dx} = \frac{x}{y \cdot \dfrac{x^2+y^2}{y^2}} = \frac{xy}{x^2+y^2}$$

Step 4 – Check the DE:
$$(x^2+y^2)\frac{dy}{dx} - xy = (x^2+y^2)\cdot\frac{xy}{x^2+y^2} - xy = xy - xy = 0$$

Conclusion: $x^2 = 2y^2\log y$ is a solution. ✓

Given: $y = ke^x - 1$

Part 1 – Verification:

Differentiate: $\dfrac{dy}{dx} = ke^x$

RHS of DE: $y + 1 = ke^x - 1 + 1 = ke^x$

Since LHS = RHS, $y = ke^x - 1$ is a solution. ✓

Part 2 – Finding $k$:

The curve passes through $(0, 1)$, so substitute $x = 0$, $y = 1$:
$$1 = ke^0 - 1 = k - 1$$
$$\Rightarrow k = 2$$

Answer: $k = \boxed{2}$

Given: $x^2 - y^2 = a^2$ … (i)

Step 1 – Differentiate w.r.t. $x$:
$$2x - 2y\frac{dy}{dx} = 0$$
$$\Rightarrow x - y\,y_1 = 0$$
$$\Rightarrow y\,y_1 = x$$

Answer: The required differential equation is $\boxed{y\,y_1 = x}$ (i.e., $y\dfrac{dy}{dx} = x$).

Given: Family of circles with centre at origin: $x^2 + y^2 = r^2$, where $r$ is an arbitrary constant.

Step 1 – Differentiate w.r.t. $x$:
$$2x + 2y\frac{dy}{dx} = 0$$
$$\Rightarrow x + y\,y_1 = 0$$

Answer: The required differential equation is $\boxed{x + y\dfrac{dy}{dx} = 0}$.

Given: A circle with centre on the $y$-axis has centre $(0, k)$ for some constant $k$. Since it passes through the origin $(0,0)$, its radius $= k$.

Equation: $x^2 + (y-k)^2 = k^2$
$$\Rightarrow x^2 + y^2 - 2ky = 0 \quad \dots (i)$$

Step 1 – Differentiate w.r.t. $x$:
$$2x + 2y\frac{dy}{dx} - 2k\frac{dy}{dx} = 0$$
$$\Rightarrow 2x + 2(y-k)y_1 = 0 \quad \dots (ii)$$

Step 2 – Eliminate $k$.

From (i): $k = \dfrac{x^2+y^2}{2y}$

Substitute into (ii):
$$2x + 2\left(y - \frac{x^2+y^2}{2y}\right)y_1 = 0$$
$$2x + 2\cdot\frac{2y^2 - x^2 - y^2}{2y}\cdot y_1 = 0$$
$$2x + \frac{(y^2-x^2)}{y}\cdot y_1 = 0$$

Multiply by $y$:
$$2xy + (y^2 - x^2)y_1 = 0$$

Answer: $\boxed{(y^2 - x^2)\dfrac{dy}{dx} + 2xy = 0}$

Given: $y = e^{2x}(a+bx)$ … (i)

Since there are two arbitrary constants, we differentiate twice.

Step 1 – First derivative:
$$y_1 = 2e^{2x}(a+bx) + be^{2x} = e^{2x}(2a+2bx+b)$$
$$\Rightarrow y_1 = 2y + be^{2x} \quad \dots (ii)$$

(using $y = e^{2x}(a+bx)$)

Step 2 – Second derivative:
$$y_2 = 2y_1 + 2be^{2x} \quad \dots (iii)$$

Step 3 – Eliminate $b$.

From (ii): $be^{2x} = y_1 - 2y$

Substitute into (iii):
$$y_2 = 2y_1 + 2(y_1 - 2y)$$
$$y_2 = 2y_1 + 2y_1 - 4y$$
$$y_2 = 4y_1 - 4y$$

Answer: $\boxed{y_2 - 4y_1 + 4y = 0}$, i.e., $\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 4y = 0$.

Given: Parabolas with vertex at origin and focus on positive $x$-axis have the form:
y^2 = 4ax, \quad a > 0 \quad \dots (i)

Step 1 – Differentiate w.r.t. $x$:
$$2y\frac{dy}{dx} = 4a \implies 4a = 2y\,y_1 \quad \dots (ii)$$

Step 2 – Substitute (ii) into (i) to eliminate $a$:
$$y^2 = (2y\,y_1)\cdot x$$
$$y^2 = 2xy\,y_1$$
$$y = 2x\,y_1$$

Answer: $\boxed{y = 2x\dfrac{dy}{dx}}$

Given: Standard ellipse with centre at origin and foci on $x$-axis:
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0 \quad \dots (i)

Two arbitrary constants $a$ and $b$ ⟹ differentiate twice.

Step 1 – Differentiate (i) w.r.t. $x$:
$$\frac{2x}{a^2} + \frac{2y\,y_1}{b^2} = 0$$
$$\Rightarrow \frac{x}{a^2} + \frac{y\,y_1}{b^2} = 0 \quad \dots (ii)$$

Step 2 – Differentiate (ii) w.r.t. $x$:
$$\frac{1}{a^2} + \frac{y_1^2 + y\,y_2}{b^2} = 0 \quad \dots (iii)$$

Step 3 – Eliminate $a^2$ and $b^2$.

From (ii): $\dfrac{1}{a^2} = -\dfrac{y\,y_1}{b^2 x}$

Substitute into (iii):
$$-\frac{y\,y_1}{b^2 x} + \frac{y_1^2 + y\,y_2}{b^2} = 0$$

Multiply by $b^2$:
$$-\frac{y\,y_1}{x} + y_1^2 + y\,y_2 = 0$$

Multiply by $x$:
$$-y\,y_1 + x\,y_1^2 + xy\,y_2 = 0$$
$$\Rightarrow xy\,y_2 + x\,y_1^2 - y\,y_1 = 0$$

Answer: $\boxed{xy\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 - y\dfrac{dy}{dx} = 0}$

Given: $\dfrac{dy}{dx} = (e^x+1)y$

Method: Separation of variables.

Step 1 – Separate variables:
$$\frac{dy}{y} = (e^x+1)\,dx$$

Step 2 – Integrate both sides:
$$\int \frac{dy}{y} = \int (e^x+1)\,dx$$
$$\ln|y| = e^x + x + C$$

Answer: $\boxed{\ln|y| = e^x + x + C}$, or equivalently $y = Ae^{e^x+x}$ where $A = e^C$.

Given: $x^5\dfrac{dy}{dx} = -y^5$

Step 1 – Separate variables:
$$\frac{dy}{y^5} = -\frac{dx}{x^5}$$
$$y^{-5}\,dy = -x^{-5}\,dx$$

Step 2 – Integrate both sides:
$$\int y^{-5}\,dy = -\int x^{-5}\,dx$$
$$\frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C$$
$$-\frac{1}{4y^4} = \frac{1}{4x^4} + C$$

Multiply by $-4$:
$$\frac{1}{y^4} = -\frac{1}{x^4} + C_1$$

Answer: $\boxed{\dfrac{1}{x^4} + \dfrac{1}{y^4} = C}$ (where $C$ is an arbitrary constant).

Given: $\dfrac{dy}{dx} = \dfrac{x+1}{2-y}$

Step 1 – Separate variables:
$$(2-y)\,dy = (x+1)\,dx$$

Step 2 – Integrate both sides:
$$\int(2-y)\,dy = \int(x+1)\,dx$$
$$2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C$$

Answer: $\boxed{2y - \dfrac{y^2}{2} = \dfrac{x^2}{2} + x + C}$

Given: $x(e^{2y}-1)\,dy + (x^2-1)e^y\,dx = 0$

Step 1 – Rearrange:
$$x(e^{2y}-1)\,dy = -(x^2-1)e^y\,dx$$
$$\frac{e^{2y}-1}{e^y}\,dy = -\frac{x^2-1}{x}\,dx$$
$$\left(e^y - e^{-y}\right)dy = -\left(x - \frac{1}{x}\right)dx$$

Step 2 – Integrate both sides:
$$\int(e^y - e^{-y})\,dy = -\int\left(x - \frac{1}{x}\right)dx$$
$$e^y + e^{-y} = -\frac{x^2}{2} + \ln|x| + C$$

Answer: $\boxed{e^y + e^{-y} = \ln|x| - \dfrac{x^2}{2} + C}$

Given: $e^x\sqrt{1-y^2}\,dx + \dfrac{y}{x}\,dy = 0$

Step 1 – Rearrange (separate variables):
$$e^x\sqrt{1-y^2}\,dx = -\frac{y}{x}\,dy$$
$$x\,e^x\,dx = -\frac{y}{\sqrt{1-y^2}}\,dy$$

Step 2 – Integrate both sides:

*LHS:* $\displaystyle\int x\,e^x\,dx = xe^x - e^x + C_1 = e^x(x-1) + C_1$ (integration by parts)

*RHS:* $\displaystyle-\int \frac{y}{\sqrt{1-y^2}}\,dy$

Let $u = 1-y^2 \Rightarrow du = -2y\,dy$:
$$-\int\frac{y}{\sqrt{u}}\cdot\frac{du}{-2y} = \frac{1}{2}\int u^{-1/2}\,du = \sqrt{u} = \sqrt{1-y^2}$$

Step 3 – Combine:
$$e^x(x-1) = \sqrt{1-y^2} + C$$

Answer: $\boxed{e^x(x-1) = \sqrt{1-y^2} + C}$

Given: $xy\dfrac{dy}{dx} = (x+2)(y+2)$, passes through $(1,-1)$.

Step 1 – Separate variables:
$$\frac{y}{y+2}\,dy = \frac{x+2}{x}\,dx$$
$$\left(1 - \frac{2}{y+2}\right)dy = \left(1 + \frac{2}{x}\right)dx$$

Step 2 – Integrate both sides:
$$y - 2\ln|y+2| = x + 2\ln|x| + C \quad \dots (i)$$

Step 3 – Apply initial condition $(1,-1)$:
$$-1 - 2\ln|1| = 1 + 2\ln|1| + C$$
$$-1 - 0 = 1 + 0 + C$$
$$C = -2$$

Step 4 – Particular solution:
$$y - 2\ln|y+2| = x + 2\ln|x| - 2$$

Answer: $\boxed{y - x + 2 = 2\ln|x| + 2\ln|y+2|}$

Given: $(x+1)\dfrac{dy}{dx} = 2xy$, $y(2) = 3$.

Step 1 – Separate variables:
$$\frac{dy}{y} = \frac{2x}{x+1}\,dx = 2\cdot\frac{x+1-1}{x+1}\,dx = 2\left(1 - \frac{1}{x+1}\right)dx$$

Step 2 – Integrate:
$$\ln|y| = 2x - 2\ln|x+1| + C$$
$$\ln|y| + 2\ln|x+1| = 2x + C$$
$$\ln|y(x+1)^2| = 2x + C$$
$$y(x+1)^2 = Ae^{2x} \quad \dots (i)$$

Step 3 – Apply $y(2) = 3$:
$$3(3)^2 = Ae^4 \implies 27 = Ae^4 \implies A = 27e^{-4}$$

Step 4 – Particular solution:
$$y(x+1)^2 = 27e^{-4}\cdot e^{2x} = 27e^{2x-4} = 27e^{2(x-2)}$$

Answer: $\boxed{y = \dfrac{27e^{2(x-2)}}{(x+1)^2}}$

Given: $\log\left(\dfrac{dy}{dx}\right) = 3x+4y$

Step 1 – Rewrite:
$$\frac{dy}{dx} = e^{3x+4y} = e^{3x}\cdot e^{4y}$$

Step 2 – Separate variables:
$$e^{-4y}\,dy = e^{3x}\,dx$$

Step 3 – Integrate:
$$\int e^{-4y}\,dy = \int e^{3x}\,dx$$
$$\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$$

Step 4 – Apply $y=0$ when $x=0$:
$$\frac{e^0}{-4} = \frac{e^0}{3} + C$$
$$-\frac{1}{4} = \frac{1}{3} + C$$
$$C = -\frac{1}{4} - \frac{1}{3} = -\frac{7}{12}$$

Step 5 – Particular solution:
$$\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12}$$

Multiply by $-12$:
$$3e^{-4y} = -4e^{3x} + 7$$

Answer: $\boxed{3e^{-4y} + 4e^{3x} = 7}$

Given: $y = y_0 e^{kt}$, $y_0 = 1$, doubling time $T = 5$ years.

Step 1: At $t = 5$, $y = 2y_0 = 2$:
$$2 = 1 \cdot e^{5k} \implies e^{5k} = 2 \implies 5k = \ln 2 \implies k = \frac{\ln 2}{5}$$

Answer: $\boxed{y = e^{\left(\frac{\ln 2}{5}\right)t}}$, i.e., $y = 2^{t/5}$.

Given: $y_0 = y(0) = 5$, growth rate $k = 2\% = 0.02$.

Answer: $\boxed{y = 5e^{0.02t}}$

Given: $y(1) = 1$, $y(10) = 100$.

Step 1: $y(1) = y_0 e^k = 1$ … (i)

$y(10) = y_0 e^{10k} = 100$ … (ii)

Step 2 – Divide (ii) by (i):
$$\frac{y_0 e^{10k}}{y_0 e^k} = \frac{100}{1} \implies e^{9k} = 100 \implies 9k = \ln 100 \implies k = \frac{\ln 100}{9} = \frac{2\ln 10}{9}$$

Step 3 – Find $y_0$ from (i):
$$y_0 = e^{-k} = e^{-\frac{2\ln 10}{9}} = 10^{-2/9}$$

Answer: $\boxed{y = 10^{-2/9}\cdot e^{\frac{2(\ln 10)}{9}t}}$, equivalently $y = 10^{(2t-2)/9}$.

Given: Principal $P_0 = 5000$, rate $r = 3\% = 0.03$, continuous compounding.

Model: $P(t) = P_0 e^{rt}$

(i) Amount after 5 years:
$$P(5) = 5000\,e^{0.03 \times 5} = 5000\,e^{0.15}$$
$$e^{0.15} \approx 1.1618$$
$$P(5) \approx 5000 \times 1.1618 = ₹5809$$

Answer (i): $\boxed{P(5) = 5000\,e^{0.15} \approx ₹5809}$

(ii) Doubling time:
$$2P_0 = P_0 e^{0.03t} \implies e^{0.03t} = 2 \implies 0.03t = \ln 2$$
$$t = \frac{\ln 2}{0.03} = \frac{0.6931}{0.03} \approx 23.1 \text{ years}$$

Answer (ii): $\boxed{t \approx 23.1 \text{ years}}$

Given: $N(t) = N_0 e^{kt}$; $N(10) = 5N_0$.

Step 1 – Find $k$:
$$5N_0 = N_0 e^{10k} \implies e^{10k} = 5 \implies k = \frac{\ln 5}{10}$$

Step 2 – Find doubling time $T$:
$$2N_0 = N_0 e^{kT} \implies kT = \ln 2 \implies T = \frac{\ln 2}{k} = \frac{10\ln 2}{\ln 5}$$
$$T = \frac{10 \times 0.6931}{1.6094} \approx \frac{6.931}{1.6094} \approx 4.31 \text{ hours}$$

Answer: $\boxed{T = \dfrac{10\ln 2}{\ln 5} \approx 4.31 \text{ hours}}$

Given: Decay model: $y(t) = y_0 e^{-0.10t}$; find $t$ when $y = \dfrac{y_0}{4}$.

Step 1:
$$\frac{y_0}{4} = y_0 e^{-0.10t}$$
$$e^{-0.10t} = \frac{1}{4} \implies -0.10t = \ln\frac{1}{4} = -\ln 4$$
$$t = \frac{\ln 4}{0.10} = \frac{2\ln 2}{0.10} = \frac{2 \times 0.6931}{0.10} \approx 13.86 \text{ years}$$

Answer: $\boxed{t = \dfrac{\ln 4}{0.1} \approx 13.86 \text{ years}}$

Given: Newton's Law of Cooling: $T(t) = T_s + (T_0 - T_s)e^{-kt}$

$T_s = 21°C$, $T_0 = 95°C$, $T(1) = 85°C$, find $t$ when $T = 51°C$.

Step 1 – Find $k$:
$$85 = 21 + (95-21)e^{-k} = 21 + 74e^{-k}$$
$$64 = 74e^{-k} \implies e^{-k} = \frac{64}{74} = \frac{32}{37}$$
$$k = \ln\frac{37}{32}$$

Step 2 – Find $t$ for $T = 51°C$:
$$51 = 21 + 74e^{-kt}$$
$$30 = 74e^{-kt} \implies e^{-kt} = \frac{30}{74} = \frac{15}{37}$$
$$-kt = \ln\frac{15}{37} \implies t = \frac{\ln(37/15)}{\ln(37/32)}$$

Step 3 – Compute:
$$\ln(37/15) = \ln 37 - \ln 15 \approx 3.6109 - 2.7081 = 0.9028$$
$$\ln(37/32) = \ln 37 - \ln 32 \approx 3.6109 - 3.4657 = 0.1452$$
$$t \approx \frac{0.9028}{0.1452} \approx 6.22 \text{ minutes}$$

Answer: $\boxed{t \approx 6.22 \text{ minutes}}$