Practice: First-Order Differential Equations

(a) Verifying $y=e^x-x-1$. Differentiate: $y^{\prime }=e^x-1$. Compute the right-hand side: $x+y=x+(e^x-x-1)=e^x-1$. The two agree identically, so the equation holds for every $x$ ✓. The initial condition: $y(0)=e^0-0-1=1-1=0$ ✓. Numerical confirmation. At $x=0,1,2$ the derivative is $0$, $1.718282$, $6.389056$ and $x+y$ is $0$, $1.718282$, $6.389056$ ✓. Verification is a differentiation, always available and independent of how the solution was found. (b) ${\displaystyle \frac{dy}{dx}}=2xy$ with $y(0)=3$. Classify. The right side factors as $(2x)(y)$, so it is separable. Rearranged as $y^{\prime }-2xy=0$ it is also linear, so either method works; separation is shorter. Note the constant solution. $g(y)=y$ vanishes at $y=0$, so $y=0$ is a solution the division will discard. Separate and integrate.

with $A=\pm e^C$. The constant entered at the integration and was absorbed during exponentiation; writing $y=e^{x^2}+C$ instead would not solve the equation. Apply the condition. $y(0)=A{e}^0=A=3$, so

Verify. $y^{\prime }=6x{e}^{x^2}$ and $2xy=2x\cdot 3e^{x^2}=6x{e}^{x^2}$ ✓. Numerically $y^{\prime }/y$ at $x=0,0.5,1$ gives $0$, $1.000000$, $2.000000$, exactly $2x$. Here $A=0$ recovers the discarded $y=0$, so nothing was permanently lost. (c) $y^{\prime }-y=e^{2x}$ with $y(0)=0$. Standard form already holds, with $P=-1$ and $Q=e^{2x}$. Integrating factor. $\mu =e^{\int (-1)dx}=e^{-x}$. Multiplying through:

and the left side is exactly ${\left(e^{-x}y\right)}^{\prime }$, which the product rule confirms as $e^{-x}{y}^{\prime }-e^{-x}y$. Integrate. $e^{-x}y=e^x+C$, so $y=e^{2x}+C{e}^x$. Apply the condition. $y(0)=1+C=0$ gives $C=-1$:

Verify. $y^{\prime }=2e^{2x}-e^x$, so $y^{\prime }-y=2e^{2x}-e^x-e^{2x}+e^x=e^{2x}$ ✓. Numerically at $x=0,0.5,1$, $y^{\prime }-y$ gives $1.000000$, $2.718282$, $7.389056$, matching $e^{2x}$ ✓. (d) Every solution of $y^{\prime }=y(1-y)$. Constant solutions first. Before dividing, find the roots of $g(y)=y(1-y)$: they are $y=0$ and $y=1$. Each gives a constant function with $y^{\prime }=0$, and $g$ vanishes there too, so both are solutions ✓. Separate. Dividing by $y(1-y)$, legitimate only where it is nonzero, and using partial fractions $\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}$:

which rearranges to

Verify. At $A=1$ and $x=0,2,5$ the derivative matches $y(1-y)$ to eight decimals; likewise at $A=4$ ✓. Which step loses them. The division by $y(1-y)$. It assumed the divisor was nonzero, which is exactly false at $y=0$ and $y=1$. The general form recovers one of them: $A=0$ gives $y=1$. But no finite $A$ gives $y=0$, since $\frac{1}{1+A{e}^{-x}}$ is never identically zero. So $y=0$ must be restored by inspection: it is a genuine solution that the method cannot produce. The complete solution set is $y={\displaystyle \frac{1}{1+A{e}^{-x}}}$ for arbitrary $A$, together with $y=0$. (e) Why a family, and why solutions can end. Why a family. The equation specifies the rate $y^{\prime }$ at each point, which determines the shape of the solution curve but not its height. Translating a solution vertically changes where it sits without changing its slope pattern, so if one curve fits, a whole family does. Formally, solving involves an integration, and every integration introduces an arbitrary constant. A first-order equation involves one integration, hence one constant, hence a one-parameter family. An initial condition supplies the missing information by naming one point the curve must pass through, which fixes the constant and selects one member. That is why the count matches: one constant, one condition. Why $y^{\prime }=y^2$ escapes. Separating gives $\int y^{-2}dy=\int dx$, so $-\frac{1}{y}=x+C$ and $y=\frac{1}{C-x}$. With $y(0)=1$, $C=1$:

Verify: at $x=0,0.5,0.9,0.99$ the derivative equals $y^2$ exactly ✓, with values $1$, $2$, $10$, $100$. As $x\to 1^{-}$ the denominator vanishes and $y\to \mathrm{\infty }$. The solution exists only on $(-\mathrm{\infty },1)$ and ceases to exist in finite $x$. Why this is not a defect in the equation. The right-hand side $y^2$ is continuous and differentiable everywhere, so the existence theorem applies at every point, but it guarantees a solution only on some interval around the starting point, not on all of $\mathbb{R}$. The blow-up is a property of the solution, produced by the feedback in the equation: the larger $y$ becomes, the faster it grows, and that acceleration is enough to reach infinity in finite time. Contrast the linear case $y^{\prime }+2y=6$, whose solution $y=3-2e^{-2x}$ exists for every real $x$. Linearity rules out this kind of runaway, which is why the linear existence theorem can promise a solution on the whole interval where $P$ and $Q$ are continuous.