Introduction to Differential Equations

(a) (i) only $\frac{dy}{dx}$ appears, so order $1$. (ii) the highest derivative is $y''$, so order $2$ — the exponent $4$ is a degree. (iii) the highest is $y'''$, so order $3$.

(b) Left side: $y'=6e^{2x}$. Right side: $2(3e^{2x}-4)+8=6e^{2x}-8+8=6e^{2x}$.

The two sides are the same function, so the equation holds for every real $x$. Comparing $Ce^{2x}-4$ with $3e^{2x}-4$ gives $C=3$, the member with $y(0)=-1$.

Separate and integrate:

$\int y\,dy=\int x\,dx \quad\Longrightarrow\quad \frac{y^2}{2}=\frac{x^2}{2}+C \quad\Longrightarrow\quad y^2=x^2+K$

$y(0)=-3$ gives $K=9$, so $y=\pm\sqrt{x^2+9}$. A solution is continuous and can never cross $y=0$, and $y(0)=-3<0$, so it stays negative: $y=-\sqrt{x^2+9}$.

Check: $y'=-\frac{x}{\sqrt{x^2+9}}=\frac{x}{y}$. Since $x^2+9>0$ for every $x$, the solution is valid on all of $\mathbb{R}$.

$h(y)=y-3$ vanishes at $y=3$, so $y\equiv3$ is a solution (both sides are $0$). For $y\neq3$:

$\int\frac{dy}{y-3}=\int 2x\,dx \quad\Longrightarrow\quad \ln|y-3|=x^2+C \quad\Longrightarrow\quad y-3=Ae^{x^2},\ A\neq0$

The sign is fixed because $y-3$ is continuous and never zero. Allowing $A=0$ puts the equilibrium back, so $y=3+Ae^{x^2}$ is every solution.

$y(0)=5$ gives $A=2$; $y(0)=3$ gives $A=0$, i.e. the equilibrium the division had discarded.

Standard form: $y'-\frac{2}{x}y=x^4$, so $p=-\frac{2}{x}$ and $\int p\,dx=-2\ln x=\ln x^{-2}$ for $x>0$, giving $\mu=x^{-2}$.

$\bigl(x^{-2}y\bigr)'=x^{-2}\cdot x^4=x^2 \quad\Longrightarrow\quad x^{-2}y=\frac{x^3}{3}+C \quad\Longrightarrow\quad y=\frac{x^5}{3}+Cx^2$

$y(1)=\frac13+C=2$ gives $C=\frac53$, so $y=\frac{x^5+5x^2}{3}$ on $(0,\infty)$. Check: $y'=\frac{5x^4+10x}{3}$, and $xy'-2y=\frac{5x^5+10x^2-2x^5-10x^2}{3}=x^5$.

(a) With $u=T-20$: $u'=ku$, so $u=75e^{kt}$ and $T=20+75e^{kt}$. Then $T(5)=70$ gives $75e^{5k}=50$, so $e^{5k}=\frac23$ and $T=20+75\left(\frac23\right)^{t/5}$.

(b) Setting $T=50$: $\left(\frac23\right)^{t/5}=\frac{30}{75}=\frac25$, so

$t=5\cdot\frac{\ln(2/5)}{\ln(2/3)}=5\cdot\frac{\ln(5/2)}{\ln(3/2)}\approx11.3\ \text{minutes}$

(c) $\left(\frac23\right)^{t/5}\to0$, so $T\to20$ — the equilibrium solution $T\equiv20$ of $T'=k(T-20)$.

(a) $y^2-1=0$ at $y=\pm1$, and both constants make each side $0$.

(b) Partial fractions and integration give

$\frac12\ln\left|\frac{y-1}{y+1}\right|=x+C \quad\Longrightarrow\quad \frac{y-1}{y+1}=Ae^{2x} \quad\Longrightarrow\quad y=\frac{1+Ae^{2x}}{1-Ae^{2x}}$

$A=0$ gives $y\equiv1$; no $A$ gives $y\equiv-1$, so that one is a singular solution.

(c) $\frac{1+A}{1-A}=0$ forces $A=-1$, so $y=\frac{1-e^{2x}}{1+e^{2x}}$. As $x\to-\infty$, $e^{2x}\to0$ and $y\to1$; dividing by $e^{2x}$ shows $y\to-1$ as $x\to\infty$.