9.4 Workshop Solutions 4

W4.1. In both cases, the region of integration is as given in the following diagram:

We have

$$\int_{1}^{2}\int_{0}^{1}(yx^{3}+y^{2}e^{x})\,dxdy=\int_{1}^{2}\Bigl[{{yx^{4}}\over{4}}+y^{2}e^{x}\Bigr]_{0}^{1}\,dy$$

$$=\int_{1}^{2}\Bigl(\Bigl[{{y}\over{4}}+y^{2}e\Bigr]-\Bigl[y^{2}\Bigr]\Bigr)dy$$

$$=\int_{1}^{2}\Bigl({{y}\over{4}}+(e-1)y^{2}\Bigr)\,dy$$

$$=\Bigl[{{y^{2}}\over{8}}+{{(e-1)y^{3}}\over{3}}\Bigr]_{1}^{2}$$

$$=\Bigl[{{4}\over{8}}+{{8(e-1)}\over{3}}\Bigr]-\Bigl[{{1}\over{8}}+{{e-1}\over{3}}\Bigr]$$

$$={{3}\over{8}}+{{7(e-1)}\over{3}}.$$

Alternatively,

$$\int_{0}^{1}\int_{1}^{2}(yx^{3}+y^{2}e^{x})\,dydx=\int_{0}^{1}\Bigl[{{y^{2}x^{3}}\over{2}}+{{y^{3}e^{2}}\over{3}}\Bigr]_{1}^{2}\,dx$$

$$=\int_{0}^{1}\Bigl(\Bigl[{{2x^{3}+{{8e^{x}}\over{3}}}}\Bigr]-\Bigl[{{x^{3}}\over{2}}+{{e^{x}}\over{3}}\Bigr]\Bigr)\,dx$$

$$=\int_{0}^{1}\Bigl({{3x^{3}}\over{2}}+{{7e^{x}}\over{3}}\Bigr)\,dx$$

$$=\Bigl[{{3x^{4}}\over{8}}+{{7e^{x}}\over{3}}\Bigr]_{0}^{1}$$

$$={{3}\over{8}}+{{7(e-1)}\over{3}}.$$

W4.2. We have $\int_{0}^{\frac{\pi}{6}}\sec^{2}(x+y)\,dx=\left[\tan(x+y)\right]_{0}^{\frac{\pi}{6}}=\tan(y+\frac{\pi}{6})-\tan y$. For the next step,

$$\int_{0}^{\frac{\pi}{6}}(\tan(y+\frac{\pi}{6})-\tan y)\,dy=\left[\log(\cos y)-\log(\cos(y+\frac{\pi}{6}))\right]_{0}^{\frac{\pi}{6}}=\log\left(\frac{\cos^{2}\frac{\pi}{6}}{\cos 0\cos\frac{\pi}{3}}\right)=\log\frac{3}{2}.$$

W4.3. a) The auxiliary equation is $s^{2}+4s+3=(s+3)(s+1)$, which has roots $-3$ and $-1$ (both single roots). Thus the general solution is $y=Ae^{-3x}+Be^{-x}$.

b) The auxiliary equation is $s^{2}+4s+4=(s+2)^{2}$, which has a double root $-2$. Thus the general solution is $y=(Ax+B)e^{-2x}$.

c) The auxiliary equation is $s^{2}+4s+5$ which has complex roots $-2\pm{\rm i}$. Therefore the general solution is $y=e^{-2x}(A\cos x+B\sin x)$.

W4.4. a) The auxiliary equation is $s^{2}-3s-4=(s-4)(s+1)$, so the general solution is $y(x)=Ae^{4x}+Be^{-x}$. For the initial value problem we have $y(0)=1$, so $A+B=1$ and hence $B=1-A$; since $y^{\prime}(x)=4Ae^{4x}-Be^{-x}$ we have $y^{\prime}(0)=4A-B$, so substituting $B=1-A$ we obtain $y^{\prime}(0)=5A-1$. Since $y^{\prime}(0)=9$ we have $A=2$ and $B=-1$, hence $y(x)=2e^{4x}-e^{-x}$.

b) Here the auxiliary equation is $s^{2}+2s+10$ which has (strictly) complex roots $-1\pm 3{\rm i}$. Hence the general solution to the equation is $y(x)=e^{-x}(A\cos 3x+B\sin 3x)$. Since $y(0)=0$, we have $A=0$, so $y(x)=Be^{-x}\sin 3x$ and we want to find the value of $B$. Now $y^{\prime}(x)=-Be^{-x}\sin 3x+3Be^{-x}\cos 3x$, so $y^{\prime}(0)=3B$. Thus $y^{\prime}(0)=1\;\Rightarrow B=\frac{1}{3}$, so the solution is $y(x)=\frac{1}{3}e^{-x}\sin 3x$.

W4.5. We divide both sides through by $(x+1)$ to obtain $\frac{dy}{dx}=\frac{x+2}{x+1}$, which we can solve by integrating. Then $\frac{x+2}{x+1}=1+\frac{1}{x+1}$, so the solution is $y=x+\log|x+1|+c$.

W4.6. This is an integrable equation, so to solve it we have to divide through by $(x^{3}+x-2)^{2}$ so that it has the form $\frac{dy}{dx}=q(x)$. Then we obtain $\frac{dy}{dx}=\frac{3x^{2}+1}{(x^{3}+x-2)^{2}}$. Fortunately we don’t have to do much work to integrate the right-hand side, since it is of the form $\frac{f^{\prime}(x)}{f(x)^{2}}$ where $f(x)=x^{3}+x-2$. Hence, after integrating, we obtain the general solution $y=\frac{-1}{x^{3}+x-2}+c$. For the particular solution with $y(0)=\frac{3}{2}$, we have $c+\frac{1}{2}=\frac{3}{2}$, so $c=1$. Hence the solution to the initial-value problem is $y=1-\frac{1}{x^{3}+x-2}$.

(*) The denominator $x^{3}+x-2$ has a real root $x=1$, and it factorizes as $(x-1)(x^{2}+x+2)$. The quadratic $x^{2}+x+2$ is irreducible (with complex roots $\frac{-1\pm\sqrt{-3}}{2}$), so $x^{3}+x-2$ is non-zero on the range $x<1$. Since we define the initial value at $x=0$, this means that the maximum possible region on which the solution is valid is the set of all $x<1$.

W4.7. To solve this equation we divide through by $y$ to obtain

$$\frac{1}{y}\frac{dy}{dx}=x^{2}+1\;\;\Rightarrow\int\frac{dy}{y}=\int(x^{2}+1)\,dx$$

and hence $\log|y|=\frac{x^{3}}{3}+x+c$. To get this into the form $y=f(x)$ we take the exponential on both sides to obtain $y=Ae^{x^{3}/3+x}$ where $A$ is a constant. Clearly $y(0)=A$, so the particular solution with $y(0)=3$ is $y=3e^{x^{3}/3+x}$.

W4.8 This is a separable differential equation, so we can solve it by multiplying through by $3(y+1)^{2}$ to obtain $\int 3(y+1)^{2}\,dy=\int x\,dx$, so that $(y+1)^{3}=\frac{x^{2}}{2}+c$. Taking cube roots, we obtain $y+1=\sqrt[3]{\frac{x^{2}}{2}+c}$ and hence the general solution to the equation is $y=-1+\sqrt[3]{\frac{x^{2}}{2}+c}$.

W4.9. We first divide through by $(x^{3}+1)$ so that the equation is in the standard form, to obtain: $\frac{dy}{dx}-\frac{3x^{2}}{x^{3}+1}y=\frac{x^{2}}{x^{3}+1}$. Now we determine the integrating factor $I(x)=e^{F(x)}$ where $F(x)=\int\frac{-3x^{2}}{x^{3}+1}\,dx$. We find $F(x)$ by substituting $u=x^{3}+1$, so that we obtain $\int\frac{-du}{u}=-\log u=-\log(x^{3}+1)$. Hence $I(x)=e^{-\log(x^{3}+1)}=\frac{1}{x^{3}+1}$. Multiplying through by $I(x)$, we obtain the integrable equation

$$\frac{1}{x^{3}+1}\frac{dy}{dx}-\frac{3x^{2}}{(x^{3}+1)^{2}}=\frac{d}{dx}\left(\frac{y}{x^{3}+1}\right)=\frac{x^{2}}{(x^{3}+1)^{2}}.$$

Integrating both sides, and observing that the right-hand side is $\frac{f^{\prime}(x)}{3f(x)}$ where $f(x)=x^{3}+1$, we find the general solution:

$$\frac{y}{x^{3}+1}=-\frac{1}{3(x^{3}+1)}+c\;\;\Rightarrow y=c(x^{3}+1)-\frac{1}{3}.$$

For the particular solution with $y(0)=1$, we have $c-\frac{1}{3}=1$ and hence $c=\frac{4}{3}$. Thus the solution to the initial-value problem is $y=\frac{4}{3}x^{3}+1$.

W4.10. We divide through by 2 and transfer the $y$ term to the left-hand side to obtain $\frac{dy}{dx}-(\tan x)y=x$. To solve this we need to multiply by the integrating factor, which is $e^{F(x)}$ where $F(x)=\int(-\tan x)\,dx$. To find $F(x)$ we observe that $\int(-\tan x)\,dx=\int\frac{-\sin x\,dx}{\cos x}$ and the derivative of denominator ($\cos x$) is the numerator ($-\sin x$), so this integral is $\log|\cos x|$. (Alternatively we make the substitution $u=\cos x$ to get the same result.) Hence the integrating factor is $I(x)=\cos x$.

Multiplying through by $I(x)$, the equation becomes $(\cos x)\frac{dy}{dx}-(\sin x)y=x\cos x$ where the left-hand side is $\frac{d}{dx}\left(y\cos x\right)$. Thus we have $y\cos x=\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+c$, where we integrated by parts in the second equality. Thus the general solution of the equation is $y=x\tan x+1+c\sec x$.