Differential Equation: Application

First Order Differential Equations using Separation of Variables

Advanced Application Problems (Separation of Variables)

The following examples illustrate applications of first order differential equations in population growth, cooling law, chemical reactions, and physics.

Example 11 (Population Growth): If population grows according to \( \frac{dP}{dt}=kP \), find \(P(t)\).

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Separate variables \[ \frac{dP}{P}=k\,dt \] Integrate \[ \int \frac{dP}{P}=\int k\,dt \] \[ \ln P = kt + C \] \[ P = Ce^{kt} \]

Example 12 (Newton's Law of Cooling): \[ \frac{dT}{dt}=-k(T-T_s) \] Find \(T(t)\).

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Separate variables \[ \frac{dT}{T-T_s}=-k\,dt \] Integrate \[ \ln |T-T_s|=-kt+C \] \[ T-T_s=Ce^{-kt} \] \[ T=T_s+Ce^{-kt} \]

Example 13 (Radioactive Decay): Solve \( \frac{dN}{dt}=-kN \)

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\[ \frac{dN}{N}=-k\,dt \] Integrate \[ \ln N=-kt+C \] \[ N=Ce^{-kt} \]

Example 14 (Chemical Reaction): Solve \[ \frac{dy}{dx}=ky(1-y) \]

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Separate variables \[ \frac{dy}{y(1-y)}=k\,dx \] Using partial fractions \[ \frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y} \] Integrate \[ \ln|y|-\ln|1-y|=kx+C \]

Example 15 (Logistic Population Model) \[ \frac{dP}{dt}=rP\left(1-\frac{P}{K}\right) \]

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Separate variables \[ \frac{dP}{P(1-P/K)}=r\,dt \] Integrate using partial fractions \[ \ln\frac{P}{K-P}=rt+C \]

Example 16: Solve \[ \frac{dy}{dx}=x\sqrt{y} \]

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Separate variables \[ \frac{dy}{\sqrt{y}}=x\,dx \] Integrate \[ 2\sqrt{y}=\frac{x^2}{2}+C \]

Example 17 (Bacterial Growth): \[ \frac{dB}{dt}=0.3B \]

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\[ \frac{dB}{B}=0.3dt \] Integrate \[ \ln B=0.3t+C \] \[ B=Ce^{0.3t} \]

Example 18: Solve \[ \frac{dy}{dx}=\frac{y}{x^2} \]

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\[ \frac{dy}{y}=\frac{dx}{x^2} \] Integrate \[ \ln y=-\frac{1}{x}+C \]

Example 19 (Evaporation Model): \[ \frac{dV}{dt}=-kV \]

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\[ \frac{dV}{V}=-kdt \] Integrate \[ \ln V=-kt+C \] \[ V=Ce^{-kt} \]

Example 20: Solve \[ \frac{dy}{dx}=\frac{x}{1+y} \]

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Separate variables \[ (1+y)dy=x\,dx \] Integrate \[ y+\frac{y^2}{2}=\frac{x^2}{2}+C \]

Example 21 (Compound Interest Model) \[ \frac{dA}{dt}=rA \]

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\[ \frac{dA}{A}=rdt \] \[ \ln A=rt+C \] \[ A=Ce^{rt} \]

Example 22: Solve \[ \frac{dy}{dx}=y^3 \]

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\[ \frac{dy}{y^3}=dx \] Integrate \[ -\frac{1}{2y^2}=x+C \]

Example 23 (Drug Elimination Model) \[ \frac{dD}{dt}=-kD \]

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\[ \frac{dD}{D}=-kdt \] \[ D=Ce^{-kt} \]

Example 24: Solve \[ \frac{dy}{dx}=\sin x \, y \]

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\[ \frac{dy}{y}=\sin x dx \] \[ \ln y=-\cos x+C \] \[ y=Ce^{-\cos x} \]

Example 25: Solve \[ \frac{dy}{dx}=y\tan x \]

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\[ \frac{dy}{y}=\tan x dx \] \[ \ln y=-\ln|\cos x|+C \] \[ y=C\sec x \]

Example 26: Solve \[ \frac{dy}{dx}=x^2y \]

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\[ \frac{dy}{y}=x^2dx \] \[ \ln y=\frac{x^3}{3}+C \]

Example 27: Solve \[ \frac{dy}{dx}=\frac{y}{\sqrt{x}} \]

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\[ \frac{dy}{y}=\frac{dx}{\sqrt{x}} \] \[ \ln y=2\sqrt{x}+C \]

Example 28 (Cooling Problem): \[ \frac{dT}{dt}=-k(T-20) \]

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\[ \frac{dT}{T-20}=-kdt \] \[ \ln|T-20|=-kt+C \]

Example 29: Solve \[ \frac{dy}{dx}=\frac{x^2}{y^2} \]

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\[ y^2dy=x^2dx \] \[ \frac{y^3}{3}=\frac{x^3}{3}+C \]

Example 30: Solve \[ \frac{dy}{dx}=e^x y \]

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\[ \frac{dy}{y}=e^x dx \] \[ \ln y=e^x+C \] \[ y=Ce^{e^x} \]