CP-9 Differential equations — coverage pack

Answer

• $e^{3x}$

Method: Here $P(x)=3$, so $I(x)=e^{\int3\,dx}=e^{3x}$. A non-zero constant multiple would be equivalent.

Answer

• $y=\dfrac{x^3}{5}+\dfrac{C}{x^2}$

Method: The integrating factor is $e^{\int2/x\,dx}=x^2$ because $x>0$. Multiplication gives $(x^2y)'=x^4$. Hence $x^2y=x^5/5+C$, and division by $x^2$ gives $y=x^3/5+C/x^2$.

Answer

• $y=\cos x+\cos^2x$
• $y(\pi/3)=\dfrac34$

Method: The integrating factor is $e^{\int2\tan x\,dx}=e^{-2\ln(\cos x)}=\sec^2x$. Thus $(y\sec^2x)'=\sin x\sec^2x$. Integration gives $y\sec^2x=\sec x+C$, so $y=\cos x+C\cos^2x$. The initial condition gives $C=1$. At $x=\pi/3$, $y=1/2+1/4=3/4$.

Answer

• General: $y=x^4+C$
• Particular: $y=x^4+2$

Method: Integrating gives $y=x^4+C$. Substituting $x=1$, $y=3$ gives $3=1+C$, so $C=2$.

Answer

• $y=5-3e^x$

Method: The auxiliary equation is $m^2-m=0$, with roots $0$ and $1$. Hence $y=A+Be^x$ and $y'=Be^x$. From $y'(0)=-3$, $B=-3$; then $y(0)=A+B=2$ gives $A=5$.

Answer

• $y=\cos(2x)-\dfrac32\sin(2x)+3x$

Method: The complementary function is $A\cos2x+B\sin2x$. A linear particular integral $px+q$ gives $4(px+q)=12x$, so $p=3$ and $q=0$. Thus $y=A\cos2x+B\sin2x+3x$. The first condition gives $A=1$. Since $y'=-2A\sin2x+2B\cos2x+3$, the second gives $2B+3=0$, so $B=-3/2$.

Answer

• $\dfrac{dT}{dt}=-k(T-18)$

Method: The excess temperature is $T-18$. Cooling means $T$ decreases when this excess is positive, so the proportionality constant must appear with a minus sign.

Answer

• $v=20(1-e^{-t/2})$
• Terminal speed $20$

Method: Rewrite as $v'+0.5v=10$. The general solution is $v=20+Ce^{-t/2}$. Since $v(0)=0$, $C=-20$, giving $v=20(1-e^{-t/2})$. As $t\to\infty$, the exponential tends to $0$, so the limiting speed is $20$ in the stated units.

Answer

• $\dfrac{dM}{dt}=1.5-\dfrac{M}{40}$
• $M=60-48e^{-t/40}$
• $t=40\ln2$ minutes
• For example, the nutrient is mixed uniformly at every instant.

Method: Nutrient enters at $3(0.5)=1.5$ grams per minute. Its concentration in the tank is $M/120$, so it leaves at $3M/120=M/40$. Thus $M'+M/40=1.5$, whose solution is $M=60+Ce^{-t/40}$. The initial mass gives $C=-48$. Setting $M=36$ gives $24=48e^{-t/40}$, so $e^{-t/40}=1/2$ and $t=40\ln2$. The derivation assumes instantaneous perfect mixing, as well as constant rates and volume.

Answer

• $y=Ae^{-2x}+Be^{-3x}$

Method: The auxiliary equation $m^2+5m+6=0$ factorises as $(m+2)(m+3)=0$. The distinct roots are $-2$ and $-3$, giving the stated two-exponential solution.

Answer

• $y=(A+Bx)e^{3x}$

Method: The auxiliary equation is $(m-3)^2=0$, so $m=3$ is a repeated root. The two independent terms are $e^{3x}$ and $xe^{3x}$.

Answer

• $y=2e^{-x}(\cos2x+\sin2x)$

Method: After division by $2$, the auxiliary equation is $m^2+2m+5=0$, with roots $-1\pm2i$. Thus $y=e^{-x}(A\cos2x+B\sin2x)$. The first condition gives $A=2$. Differentiating and setting $x=0$ gives $y'(0)=-A+2B=2$, so $B=2$.

Answer

• $y=Ae^{2x}+Be^{-x}-3$

Method: The auxiliary equation $(m-2)(m+1)=0$ gives $y_{\mathrm c}=Ae^{2x}+Be^{-x}$. For a constant PI $K$, substitution gives $-2K=6$, so $K=-3$.

Answer

• $y=Ae^{2x}+Be^{-2x}+2xe^{2x}$

Method: The complementary function is $Ae^{2x}+Be^{-2x}$. Since $e^{2x}$ is already present, try $y_{\mathrm p}=Cxe^{2x}$. Then $y_{\mathrm p}''-4y_{\mathrm p}=4Ce^{2x}$, so $C=2$.

Answer

• $y=-e^{-x}(\cos2x+\sin2x)+2\cos x+\sin x$

Method: The auxiliary roots are $-1\pm2i$, so $y_{\mathrm c}=e^{-x}(A\cos2x+B\sin2x)$. Try $y_{\mathrm p}=a\cos x+b\sin x$. Substitution gives $4a+2b=10$ and $-2a+4b=0$, so $a=2$, $b=1$. From $y(0)=1$, $A+2=1$ and $A=-1$. The derivative at $0$ is $-A+2B+1$; setting this to $0$ gives $B=-1$. Combining the parts yields the stated solution.

Answer

• The discriminant is negative; $y=e^{-2x}(A\cos2x+B\sin2x)$.

Method: The discriminant is $4^2-4(8)=-16<0$. The roots are $-2\pm2i$, which give the damped sine-cosine form.

Answer

• $k=8$; $y=(A+Bx)e^{-4x}$

Method: A repeated root requires discriminant $k^2-64=0$. Since $k>0$, $k=8$. The repeated root is $-k/2=-4$, so the solution is $(A+Bx)e^{-4x}$.

Answer

• Distinct real roots for $p<-7/2$, a repeated root for $p=-7/2$, and complex conjugate roots for $p>-7/2$.
• At $p=-7/2$, $y=(A+Bx)e^{5x/2}$.

Method: The discriminant is $[2(p+1)]^2-4(p^2+4p+8)=-4(2p+7)$. It is positive, zero or negative according as $p<-7/2$, $p=-7/2$ or $p>-7/2$. At equality the repeated root is $-[2(p+1)]/2=-(p+1)=5/2$, giving $(A+Bx)e^{5x/2}$.

Answer

• $x=A\cos3t+B\sin3t$; period $2\pi/3$.

Method: Comparison with $x''=-\omega^2x$ gives $\omega=3$. Substitute this angular frequency into the general SHM solution and use $T=2\pi/\omega$.

Answer

• $x=4\cos2t+3\sin2t$
• Amplitude $5$, period $\pi$, maximum speed $10$.

Method: Here $\omega=2$, so $x=A\cos2t+B\sin2t$. The conditions give $A=4$ and $2B=6$, hence $B=3$. The amplitude is $\sqrt{4^2+3^2}=5$, the period is $2\pi/2=\pi$, and the maximum speed is $\omega R=2(5)=10$.

Answer

• $x=\cos(4t+\pi/6)$
• $t=\pi/12$
• $x''=8$ when $x=-1/2$.

Method: In $x=A\cos4t+B\sin4t$, the conditions give $A=\sqrt3/2$ and $4B=-2$, so $B=-1/2$. Thus $R=1$ and $x=\cos(4t+\pi/6)$. The first negative-direction equilibrium crossing has phase $\pi/2$, so $4t+\pi/6=\pi/2$ and $t=\pi/12$. Finally $x''=-16x=-16(-1/2)=8$.

Answer

• Underdamped; $x=e^{-3t}(A\cos4t+B\sin4t)$.

Method: The auxiliary equation has roots $-3\pm4i$. The non-zero imaginary part produces oscillation, while the negative real part gives the decaying envelope $e^{-3t}$.

Answer

• $c=6$
• $x=2(1+t)e^{-3t}$
• The displacement tends to $0$ without oscillation.

Method: Critical damping requires $c^2-36=0$, so positive $c$ gives $c=6$. The repeated root is $-3$, hence $x=(A+Bt)e^{-3t}$. From $x(0)=2$, $A=2$; from $x'(0)=B-3A=-4$, $B=2$. Both the exponential and polynomial-exponential term tend to $0$, and the repeated real root means no oscillation.

Answer

• $x=e^{-2t}(-\dfrac12\cos4t-\dfrac38\sin4t)+\dfrac12\cos2t+\dfrac14\sin2t$
• The transient term decays, leaving a steady oscillation of angular frequency $2$ and amplitude $\sqrt5/4$.

Method: The auxiliary roots are $-2\pm4i$, so $x_{\mathrm c}=e^{-2t}(A\cos4t+B\sin4t)$. Try $x_{\mathrm p}=a\cos2t+b\sin2t$. Substitution gives $16a+8b=10$ and $-8a+16b=0$, hence $a=1/2$ and $b=1/4$. From $x(0)=0$, $A=-1/2$. At $t=0$, $x'=-2A+4B+2b=0$, so $B=-3/8$. The complementary term tends to $0$, while the PI remains with amplitude $\sqrt{(1/2)^2+(1/4)^2}=\sqrt5/4$ and angular frequency $2$.

Answer

• $u''-2u'-3u=0$

Method: Differentiate $u'=2u+v$ to get $u''=2u'+v'$. Substitute $v'=3u$, then rearrange to $u''-2u'-3u=0$.

Answer

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

Method: From the first equation, $y=\tfrac12(x'-x)$, so $y'=\tfrac12(x''-x')$. Substituting into $y'=2x+y$ gives $\tfrac12(x''-x')=2x+\tfrac12(x'-x)$, i.e. $x''-2x'-3x=0$. The auxiliary equation $m^2-2m-3=0$ has roots $m=3$ and $m=-1$, so $x=Ae^{3t}+Be^{-t}$. Then $y=\tfrac12(x'-x)=Ae^{3t}-Be^{-t}$. The conditions $x(0)=3$, $y(0)=1$ give $A+B=3$ and $A-B=1$, so $A=2$, $B=1$: $x=2e^{3t}+e^{-t}$ and $y=2e^{3t}-e^{-t}$.

Answer

• $A=3e^{-t}+e^{-4t}$
• $B=3e^{-t}-2e^{-4t}$
• $B$ is greatest at $t=\dfrac13\ln(8/3)$, with maximum $\dfrac94(3/8)^{1/3}$.
• Both amounts tend to $0$ as $t\to\infty$.

Method: From $A'=-2A+B$, $B=A'+2A$. Differentiate and use the second equation to obtain $A''+5A'+4A=0$. Thus $A=Ce^{-t}+De^{-4t}$. Now $A(0)=4$ and $A'(0)=-2(4)+1=-7$, giving $C=3$, $D=1$. Hence $B=A'+2A=3e^{-t}-2e^{-4t}$. For a maximum, $B'=-3e^{-t}+8e^{-4t}=0$, so $e^{3t}=8/3$ and $t=(1/3)\ln(8/3)$. Writing $r=e^{-t}=(3/8)^{1/3}$ gives $e^{-4t}=(3/8)r$, hence $B_{\max}=3r-2(3/8)r=(9/4)r$. Both exponential terms vanish in the long term.