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Differential Equations and Modeling

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Differential Equations and Modeling — CBSE Class 12 Applied Mathematics.

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Exercise 1 — Order and Degree of Differential Equations

1Determine the order and degree (if defined) of the differential equation: xdydx+2y=x2,x≠0x \frac{dy}{dx} + 2y = x^2, \quad x \neq 0Show solution
Given: xdydx+2y=x2x \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 dydx\dfrac{dy}{dx} (first derivative).
- It appears with power 1.

Answer: Order = 1, Degree = 1

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2Determine the order and degree (if defined) of the differential equation: dydx+ey=0\frac{dy}{dx} + e^y = 0Show solution
Given: dydx+ey=0\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 dydx\dfrac{dy}{dx} (first derivative), appearing with power 1.
- The term eye^y involves yy (not a derivative), so it does not affect the degree.

Answer: Order = 1, Degree = 1

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3Determine the order and degree (if defined) of the differential equation: d2ydx2+dydx−6y=0\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0Show solution
Given: d2ydx2+dydx−6y=0\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - 6y = 0

Working:
- The highest-order derivative is d2ydx2\dfrac{d^2y}{dx^2} (second derivative).
- It appears with power 1.

Answer: Order = 2, Degree = 1

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4Determine the order and degree (if defined) of the differential equation: (dydx)4+3y(d3ydx3)=0\left(\frac{dy}{dx}\right)^4 + 3y\left(\frac{d^3y}{dx^3}\right) = 0Show solution
Given: (dydx)4+3y(d3ydx3)=0\left(\dfrac{dy}{dx}\right)^4 + 3y\left(\dfrac{d^3y}{dx^3}\right) = 0

Working:
- Derivatives present: dydx\dfrac{dy}{dx} (order 1) and d3ydx3\dfrac{d^3y}{dx^3} (order 3).
- Highest-order derivative: d3ydx3\dfrac{d^3y}{dx^3} — this is order 3.
- The highest-order derivative d3ydx3\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 d2ydx2\dfrac{d^2y}{dx^2}, the answer would be order 2, degree 1. Students should follow the equation as printed.)*

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5Determine the order and degree (if defined) of the differential equation: (y′′′)2+(y′′)3+(y′)4+y5=0(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0, where y′=dydxy' = \frac{dy}{dx}, y′′=d2ydx2y'' = \frac{d^2y}{dx^2} (as given, though the problem states y′′=d3ydx3y'' = \frac{d^3y}{dx^3} and y′′′=d4ydx4y''' = \frac{d^4y}{dx^4}; we use the standard notation y′′′y''' = third derivative).Show solution
Given: (y′′′)2+(y′′)3+(y′)4+y5=0(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0

As stated in the problem: y′=dydxy' = \dfrac{dy}{dx}, y′′=d2ydx2y'' = \dfrac{d^2y}{dx^2}, y′′′=d3ydx3y''' = \dfrac{d^3y}{dx^3}.

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

Answer: Order = 3, Degree = 2

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Exercise 2 — Verification of Solutions

1Verify that y=ae−xy = ae^{-x} is a solution of dydx+y=0\frac{dy}{dx} + y = 0.Show solution
Given: y=ae−xy = ae^{-x}

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

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

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

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2Verify that y=1+x2y = \sqrt{1+x^2} is a solution of dydx=xy1+x2\frac{dy}{dx} = \frac{xy}{1+x^2}.Show solution
Given: y=1+x2=(1+x2)1/2y = \sqrt{1+x^2} = (1+x^2)^{1/2}

Step 1 – Differentiate:
dydx=12(1+x2)−1/2⋅2x=x1+x2\frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1+x^2}}

Step 2 – Compute RHS:
xy1+x2=x⋅1+x21+x2=x1+x2\frac{xy}{1+x^2} = \frac{x \cdot \sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}

Step 3 – Compare:
LHS=x1+x2=RHS\text{LHS} = \frac{x}{\sqrt{1+x^2}} = \text{RHS}

Conclusion: y=1+x2y = \sqrt{1+x^2} is a solution. ✓

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3Verify that xy=log⁡y+cxy = \log y + c is a solution of dydx=y21−xy\frac{dy}{dx} = \frac{y^2}{1-xy}, (xy≠1)(xy \neq 1).Show solution
Given: xy=log⁡y+cxy = \log y + c

**Step 1 – Differentiate implicitly with respect to xx:**
y+xdydx=1ydydxy + x\frac{dy}{dx} = \frac{1}{y}\frac{dy}{dx}

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

**Step 3 – Solve for dydx\dfrac{dy}{dx}:**
dydx=y⋅y1−xy=y21−xy\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+cxy = \log y + c is a solution. ✓

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4Verify that ax2+by2=1ax^2 + by^2 = 1 is a solution of x(yy2+y12)=yy1x(yy_2 + y_1^2) = yy_1, where y1=dydxy_1 = \frac{dy}{dx}, y2=d2ydx2y_2 = \frac{d^2y}{dx^2}.Show solution
Given: ax2+by2=1ax^2 + by^2 = 1 … (i)

**Step 1 – First differentiation (w.r.t. xx):**
2ax+2by y1=0  ⟹  ax+by y1=0…(ii)2ax + 2by\,y_1 = 0 \implies ax + by\,y_1 = 0 \quad \dots (ii)

Step 2 – Second differentiation:
a+b(y12+y y2)=0…(iii)a + b(y_1^2 + y\,y_2) = 0 \quad \dots (iii)

**Step 3 – Eliminate aa and bb.**

From (ii): a=−by y1xa = -\dfrac{by\,y_1}{x}

Substitute into (iii):
−by y1x+b(y12+y y2)=0-\frac{by\,y_1}{x} + b(y_1^2 + y\,y_2) = 0

Divide by bb (assuming b≠0b \neq 0):
−y y1x+y12+y y2=0-\frac{y\,y_1}{x} + y_1^2 + y\,y_2 = 0

Multiply throughout by xx:
−y y1+x y12+xy y2=0-y\,y_1 + x\,y_1^2 + xy\,y_2 = 0
⇒x(y y2+y12)=y y1\Rightarrow x(y\,y_2 + y_1^2) = y\,y_1

Conclusion: This is exactly the given DE. Hence ax2+by2=1ax^2 + by^2 = 1 is a solution. ✓

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5Verify that y=(a+bx)e2xy = (a+bx)e^{2x} is a solution of y2−4y1+4y=0y_2 - 4y_1 + 4y = 0.Show solution
Given: y=(a+bx)e2xy = (a+bx)e^{2x}

Step 1 – First derivative:
y1=b e2x+2(a+bx)e2x=e2x[b+2(a+bx)]=e2x(2a+b+2bx)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:
y2=2e2x(2a+b+2bx)+e2x(2b)=e2x(4a+2b+4bx+2b)=e2x(4a+4b+4bx)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 y2−4y1+4yy_2 - 4y_1 + 4y:**
y2−4y1+4y=e2x(4a+4b+4bx)−4e2x(2a+b+2bx)+4e2x(a+bx)y_2 - 4y_1 + 4y = e^{2x}(4a+4b+4bx) - 4e^{2x}(2a+b+2bx) + 4e^{2x}(a+bx)
=e2x[(4a+4b+4bx)−(8a+4b+8bx)+(4a+4bx)]= e^{2x}\left[(4a+4b+4bx) - (8a+4b+8bx) + (4a+4bx)\right]
=e2x[4a+4b+4bx−8a−4b−8bx+4a+4bx]= e^{2x}\left[4a+4b+4bx - 8a-4b-8bx+4a+4bx\right]
=e2x[(4a−8a+4a)+(4b−4b)+(4bx−8bx+4bx)]= e^{2x}\left[(4a-8a+4a)+(4b-4b)+(4bx-8bx+4bx)\right]
=e2x[0+0+0]=0= e^{2x}[0+0+0] = 0

Conclusion: y=(a+bx)e2xy = (a+bx)e^{2x} is a solution of y2−4y1+4y=0y_2 - 4y_1 + 4y = 0. ✓

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6Verify that x2=2y2log⁡yx^2 = 2y^2 \log y is a solution of (x2+y2)dydx−xy=0(x^2+y^2)\frac{dy}{dx} - xy = 0.Show solution
Given: x2=2y2log⁡yx^2 = 2y^2 \log y … (i)

**Step 1 – Differentiate implicitly w.r.t. xx:**
2x=2[2ylog⁡y⋅dydx+y2⋅1y⋅dydx]2x = 2\left[2y\log y \cdot \frac{dy}{dx} + y^2 \cdot \frac{1}{y}\cdot\frac{dy}{dx}\right]
2x=2dydx[2ylog⁡y+y]2x = 2\frac{dy}{dx}\left[2y\log y + y\right]
x=dydx⋅y(2log⁡y+1)x = \frac{dy}{dx}\cdot y(2\log y + 1)

⇒dydx=xy(2log⁡y+1)…(ii)\Rightarrow \frac{dy}{dx} = \frac{x}{y(2\log y+1)} \quad \dots (ii)

Step 2 – From (i): 2log⁡y=x2y22\log y = \dfrac{x^2}{y^2}, so 2log⁡y+1=x2+y2y22\log y + 1 = \dfrac{x^2+y^2}{y^2}.

Step 3 – Substitute into (ii):
dydx=xy⋅x2+y2y2=xyx2+y2\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:
(x2+y2)dydx−xy=(x2+y2)⋅xyx2+y2−xy=xy−xy=0(x^2+y^2)\frac{dy}{dx} - xy = (x^2+y^2)\cdot\frac{xy}{x^2+y^2} - xy = xy - xy = 0

Conclusion: x2=2y2log⁡yx^2 = 2y^2\log y is a solution. ✓

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7Verify that y=kex−1y = ke^x - 1 is a solution of dydx=y+1\frac{dy}{dx} = y+1. Also find kk so that the solution curve passes through (0,1)(0,1).Show solution
Given: y=kex−1y = ke^x - 1

Part 1 – Verification:

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

RHS of DE: y+1=kex−1+1=kexy + 1 = ke^x - 1 + 1 = ke^x

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

**Part 2 – Finding kk:**

The curve passes through (0,1)(0, 1), so substitute x=0x = 0, y=1y = 1:
1=ke0−1=k−11 = ke^0 - 1 = k - 1
⇒k=2\Rightarrow k = 2

Answer: k=2k = \boxed{2}

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Exercise 3 — Formation of Differential Equations

1Form the differential equation not containing the arbitrary constant and satisfied by x2−y2=a2x^2 - y^2 = a^2, where aa is an arbitrary constant.Show solution
Given: x2−y2=a2x^2 - y^2 = a^2 … (i)

**Step 1 – Differentiate w.r.t. xx:**
2x−2ydydx=02x - 2y\frac{dy}{dx} = 0
⇒x−y y1=0\Rightarrow x - y\,y_1 = 0
⇒y y1=x\Rightarrow y\,y_1 = x

Answer: The required differential equation is y y1=x\boxed{y\,y_1 = x} (i.e., ydydx=xy\dfrac{dy}{dx} = x).

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2Find the differential equation of the family of circles having centre at the origin.Show solution
Given: Family of circles with centre at origin: x2+y2=r2x^2 + y^2 = r^2, where rr is an arbitrary constant.

**Step 1 – Differentiate w.r.t. xx:**
2x+2ydydx=02x + 2y\frac{dy}{dx} = 0
⇒x+y y1=0\Rightarrow x + y\,y_1 = 0

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

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3Form the differential equation of the family of circles having centre on the yy-axis and passing through the origin.Show solution
Given: A circle with centre on the yy-axis has centre (0,k)(0, k) for some constant kk. Since it passes through the origin (0,0)(0,0), its radius =k= k.

Equation: x2+(y−k)2=k2x^2 + (y-k)^2 = k^2
⇒x2+y2−2ky=0…(i)\Rightarrow x^2 + y^2 - 2ky = 0 \quad \dots (i)

**Step 1 – Differentiate w.r.t. xx:**
2x+2ydydx−2kdydx=02x + 2y\frac{dy}{dx} - 2k\frac{dy}{dx} = 0
⇒2x+2(y−k)y1=0…(ii)\Rightarrow 2x + 2(y-k)y_1 = 0 \quad \dots (ii)

**Step 2 – Eliminate kk.**

From (i): k=x2+y22yk = \dfrac{x^2+y^2}{2y}

Substitute into (ii):
2x+2(y−x2+y22y)y1=02x + 2\left(y - \frac{x^2+y^2}{2y}\right)y_1 = 0
2x+2⋅2y2−x2−y22y⋅y1=02x + 2\cdot\frac{2y^2 - x^2 - y^2}{2y}\cdot y_1 = 0
2x+(y2−x2)y⋅y1=02x + \frac{(y^2-x^2)}{y}\cdot y_1 = 0

Multiply by yy:
2xy+(y2−x2)y1=02xy + (y^2 - x^2)y_1 = 0

Answer: (y2−x2)dydx+2xy=0\boxed{(y^2 - x^2)\dfrac{dy}{dx} + 2xy = 0}

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4Form the differential equation representing the family of curves y=e2x(a+bx)y = e^{2x}(a+bx), where a,ba, b are arbitrary constants.Show solution
Given: y=e2x(a+bx)y = e^{2x}(a+bx) … (i)

Since there are two arbitrary constants, we differentiate twice.

Step 1 – First derivative:
y1=2e2x(a+bx)+be2x=e2x(2a+2bx+b)y_1 = 2e^{2x}(a+bx) + be^{2x} = e^{2x}(2a+2bx+b)
⇒y1=2y+be2x…(ii)\Rightarrow y_1 = 2y + be^{2x} \quad \dots (ii)

(using y=e2x(a+bx)y = e^{2x}(a+bx))

Step 2 – Second derivative:
y2=2y1+2be2x…(iii)y_2 = 2y_1 + 2be^{2x} \quad \dots (iii)

**Step 3 – Eliminate bb.**

From (ii): be2x=y1−2ybe^{2x} = y_1 - 2y

Substitute into (iii):
y2=2y1+2(y1−2y)y_2 = 2y_1 + 2(y_1 - 2y)
y2=2y1+2y1−4yy_2 = 2y_1 + 2y_1 - 4y
y2=4y1−4yy_2 = 4y_1 - 4y

Answer: y2−4y1+4y=0\boxed{y_2 - 4y_1 + 4y = 0}, i.e., d2ydx2−4dydx+4y=0\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 4y = 0.

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5Find the differential equation representing the parabolas having their vertices at the origin and foci on the positive direction of the xx-axis.Show solution
Given: Parabolas with vertex at origin and focus on positive xx-axis have the form:
y2=4ax,a>0…(i)y^2 = 4ax, \quad a > 0 \quad \dots (i)

**Step 1 – Differentiate w.r.t. xx:**
2ydydx=4a  ⟹  4a=2y y1…(ii)2y\frac{dy}{dx} = 4a \implies 4a = 2y\,y_1 \quad \dots (ii)

**Step 2 – Substitute (ii) into (i) to eliminate aa:**
y2=(2y y1)⋅xy^2 = (2y\,y_1)\cdot x
y2=2xy y1y^2 = 2xy\,y_1
y=2x y1y = 2x\,y_1

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

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6Form the differential equation of the family of ellipses having their foci on the xx-axis and centre at the origin.Show solution
Given: Standard ellipse with centre at origin and foci on xx-axis:
x2a2+y2b2=1,a>b>0…(i)\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0 \quad \dots (i)

Two arbitrary constants aa and bb ⟹ differentiate twice.

**Step 1 – Differentiate (i) w.r.t. xx:**
2xa2+2y y1b2=0\frac{2x}{a^2} + \frac{2y\,y_1}{b^2} = 0
⇒xa2+y y1b2=0…(ii)\Rightarrow \frac{x}{a^2} + \frac{y\,y_1}{b^2} = 0 \quad \dots (ii)

**Step 2 – Differentiate (ii) w.r.t. xx:**
1a2+y12+y y2b2=0…(iii)\frac{1}{a^2} + \frac{y_1^2 + y\,y_2}{b^2} = 0 \quad \dots (iii)

**Step 3 – Eliminate a2a^2 and b2b^2.**

From (ii): 1a2=−y y1b2x\dfrac{1}{a^2} = -\dfrac{y\,y_1}{b^2 x}

Substitute into (iii):
−y y1b2x+y12+y y2b2=0-\frac{y\,y_1}{b^2 x} + \frac{y_1^2 + y\,y_2}{b^2} = 0

Multiply by b2b^2:
−y y1x+y12+y y2=0-\frac{y\,y_1}{x} + y_1^2 + y\,y_2 = 0

Multiply by xx:
−y y1+x y12+xy y2=0-y\,y_1 + x\,y_1^2 + xy\,y_2 = 0
⇒xy y2+x y12−y y1=0\Rightarrow xy\,y_2 + x\,y_1^2 - y\,y_1 = 0

Answer: xyd2ydx2+x(dydx)2−ydydx=0\boxed{xy\dfrac{d^2y}{dx^2} + x\left(\dfrac{dy}{dx}\right)^2 - y\dfrac{dy}{dx} = 0}

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Exercise 4 — Solving Differential Equations

1Find the general solution of dydx=(ex+1)y\frac{dy}{dx} = (e^x+1)y.Show solution
Given: dydx=(ex+1)y\dfrac{dy}{dx} = (e^x+1)y

Method: Separation of variables.

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

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

Answer: ln⁡∣y∣=ex+x+C\boxed{\ln|y| = e^x + x + C}, or equivalently y=Aeex+xy = Ae^{e^x+x} where A=eCA = e^C.

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2Find the general solution of x5dydx=−y5x^5 \frac{dy}{dx} = -y^5.Show solution
Given: x5dydx=−y5x^5\dfrac{dy}{dx} = -y^5

Step 1 – Separate variables:
dyy5=−dxx5\frac{dy}{y^5} = -\frac{dx}{x^5}
y−5 dy=−x−5 dxy^{-5}\,dy = -x^{-5}\,dx

Step 2 – Integrate both sides:
∫y−5 dy=−∫x−5 dx\int y^{-5}\,dy = -\int x^{-5}\,dx
y−4−4=−x−4−4+C\frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C
−14y4=14x4+C-\frac{1}{4y^4} = \frac{1}{4x^4} + C

Multiply by −4-4:
1y4=−1x4+C1\frac{1}{y^4} = -\frac{1}{x^4} + C_1

Answer: 1x4+1y4=C\boxed{\dfrac{1}{x^4} + \dfrac{1}{y^4} = C} (where CC is an arbitrary constant).

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3Find the general solution of dydx=x+12−y\frac{dy}{dx} = \frac{x+1}{2-y}.
4Find the general solution of x(e2y−1) dy+(x2−1)ey dx=0x(e^{2y}-1)\,dy + (x^2-1)e^y\,dx = 0.
5Find the general solution of ex1−y2 dx+yx dy=0e^x\sqrt{1-y^2}\,dx + \frac{y}{x}\,dy = 0.
6Find the equation of the curve passing through the point (1,−1)(1,-1) whose differential equation is xydydx=(x+2)(y+2)xy\dfrac{dy}{dx} = (x+2)(y+2).
7Solve (x+1)dydx=2xy(x+1)\frac{dy}{dx} = 2xy, given that y(2)=3y(2) = 3.
8Find the particular solution of log⁡(dydx)=3x+4y\log\left(\frac{dy}{dx}\right) = 3x+4y, given that y=0y = 0 when x=0x = 0.

Exercise 5 — Differential Equations and Mathematical Modeling

1(i)Find an exponential growth model y=y0ekty = y_0 e^{kt} satisfying: y0=1y_0 = 1 and doubling time t=5t = 5 years.
1(ii)Find an exponential growth model y=y0ekty = y_0 e^{kt} satisfying: y(0)=5y(0) = 5 and growth rate =2%= 2\%.
1(iii)Find an exponential growth model y=y0ekty = y_0 e^{kt} satisfying: y(1)=1y(1) = 1 and y(10)=100y(10) = 100.
2Gaurav deposited ₹5000 in an account paying 3% interest compounded continuously for 5 years. (i) Find the total amount at the end of 5 years. (ii) How long will it take for the money to double?
3In a certain culture of bacteria, the number of bacteria increased 5 times in 10 hours. How long did it take for the number of bacteria to double?
4The amount of oil pumped from one of the wells decreases at the continuous rate of 10% per year. When will the well's output fall to one-fourth of its present value?
5A cup of tea with temperature 95°C95°C is placed in a room with constant temperature 21°C21°C. How many minutes will it take to reach 51°C51°C if it cools to 85°C85°C in 1 minute?
6A cake is removed from an oven at 250°F250°F and left to cool at room temperature 70°F70°F. After 30 minutes the temperature is 150°F150°F. After how much time will it be 100°F100°F?
7Radium decomposes at a rate proportional to the amount present. If half the original amount disappears in 1600 years, find the percentage lost in 100 years.
8Half-life of radioactive carbon-14 is 5700 years. A certain bone was observed to contain 75% of carbon-14 as compared to what is present in living creatures. Determine its antiquity.
9If 600 grams of a radioactive substance are present initially and 3 years later only 300 grams remain. How much of the substance will be present after 6 years?
10The output of a radioactive power supply is y=50e−0.004ty = 50e^{-0.004t} watts, where tt is in days. (a) How much power will be available at the end of 90 days? (b) How long will it take for the power to be half of its original strength?
11Use the exponential growth model to show that the time it takes for a population to double (from AA to 2A2A) is given by t=ln⁡2kt = \frac{\ln 2}{k}.

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