First-Order Differential Equations: Basics, Linear Forms & Applications

The order of a differential equation is defined as the order of the highest derivative that appears in the equation. In this equation, we have two derivatives: dy/dx (first derivative, order 1) and d²y/dx² (second derivative, order 2). The highest order derivative present is the second derivative d²y/dx². Therefore, the order of this differential equation is 2, making it a second order differential equation. The coefficients (like x³) and the powers of the dependent variable (like y⁴) do not determine the order; only the highest derivative matters.

To find solutions to dy/dx = 3x², we integrate both sides with respect to x. The integration gives us y = ∫3x²dx = 3(x³/3) + C = x³ + C, where C is the constant of integration. This constant C can take any real number value. For each different value of C, we obtain a different solution function. For example, when C = 0, we get y = x³. When C = 5, we get y = x³ + 5. Since there are infinitely many possible real values for C, there are infinitely many distinct solution functions to this differential equation. This illustrates that first order differential equations typically have families of solutions rather than a single unique solution.

To find the general solution, we need to integrate the right-hand side of the differential equation with respect to x. We compute the antiderivative of 5x⁴ - 2x + 7 term by term. For the first term, ∫5x⁴dx = 5(x⁵/5) = x⁵. For the second term, ∫(-2x)dx = -2(x²/2) = -x². For the third term, ∫7dx = 7x. When we combine these results and add the constant of integration C, we obtain y = x⁵ - x² + 7x + C. This is the general solution because it contains the arbitrary constant C, which means it represents all possible solutions to the differential equation. We can verify this by differentiating: d/dx(x⁵ - x² + 7x + C) = 5x⁴ - 2x + 7, which matches the original equation.

First, we find the general solution by integrating both sides of dy/dx = 2x. Integrating gives us y = ∫2xdx = x² + C, where C is the constant of integration. Next, we use the initial condition y(2) = 9 to determine the specific value of C. Substituting x = 2 and y = 9 into the general solution: 9 = (2)² + C, which simplifies to 9 = 4 + C. Solving for C, we subtract 4 from both sides: C = 9 - 4 = 5. Now we substitute this value of C back into the general solution to obtain the particular solution: y = x² + 5. We can verify this solution by checking both the differential equation (dy/dx = 2x) and the initial condition (when x = 2, y = 2² + 5 = 9).

To solve this problem, we first find the general solution by integrating the right-hand side. The integral of e^(3x) with respect to x is (⅓)e^(3x) + C, because the derivative of (⅓)e^(3x) is (⅓) × 3e^(3x) = e^(3x). So the general solution is y = (⅓)e^(3x) + C. Next, we use the point (0, 4) to find the value of C. Substituting x = 0 and y = 4 into the equation: 4 = (⅓)e^(3×0) + C. Since e⁰= 1, this becomes 4 = (⅓)(1) + C, which simplifies to 4 = 1/3 + C. To solve for C, subtract 1/3 from both sides: C = 4 - 1/3 = 12/3 - 1/3 = 11/3. Therefore, the particular solution is y = (⅓)e^(3x) + 11/3.

This is a separable differential equation of the form dy/dx = g(y). We can solve it by separating variables. First, we rewrite the equation as (1/y)dy = -4dx. Now we integrate both sides: ∫(1/y)dy = ∫(-4)dx. The left integral gives ln|y|, and the right integral gives -4x + C. So we have ln|y| = -4x + C. To solve for y, we exponentiate both sides: e^(ln|y|) = e^(-4x + C), which gives |y| = e^C × e^(-4x). Since e^C is a positive constant, we can write this as y = ±e^C × e^(-4x). The ±e^C can be replaced by a new constant K, so y = Ke^(-4x). Now we apply the initial condition y(0) = 3: 3 = Ke^(-4×0) = Ke⁰= K×1 = K. Therefore, K = 3. Substituting back, we get the particular solution y = 3e^(-4x).

Solving dy/dx = -0.5y, we separate variables: dy/y = -0.5dx. Integrating gives ln|y| = -0.5x + C. Exponentiating: y = Ke^(-0.5x). The constant K is determined by the initial condition: at x = 0, y(0) = K·e⁰= K, so K = y₀. Thus the solution is y = y₀·e^(-0.5x). As x → ∞, the exponential term e^(-0.5x) → 0, so y → 0 for any initial y₀. However, at any finite x, the value of y is proportional to y₀. For example, at x = 2, y = y₀·e^(-1). If y₀ = 10, y(2) ≈ 3.68. If y₀ = 20, y(2) ≈ 7.36, which is exactly double. Therefore, curves with larger initial values remain proportionally larger at every x while all approaching zero asymptotically. This creates a family of curves that all converge to zero but maintain their relative ordering.

For a linear differential equation of the form dy/dx + P(x)y = Q(x), we use an integrating factor μ(x) = e^(∫P(x)dx). Here, P(x) = 3, so ∫P(x)dx = ∫3dx = 3x. The integrating factor is μ(x) = e^(3x). Multiply the entire equation by this factor: e^(3x)dy/dx + 3e^(3x)y = 6e^(3x). The left side is now the derivative of μ(x)y: d/dx(e^(3x)y) = 6e^(3x). Integrate both sides: ∫d/dx(e^(3x)y)dx = ∫6e^(3x)dx. The left side simplifies to e^(3x)y. For the right side, ∫6e^(3x)dx = 6(e^(3x)/3) + C = 2e^(3x) + C. So we have e^(3x)y = 2e^(3x) + C. Divide both sides by e^(3x): y = 2 + Ce^(-3x). Note that options A and C are mathematically equivalent (just written in different order), but the standard form writes the particular solution first: y = 2 + Ce^(-3x).

This is a linear differential equation in the form dy/dx + P(x)y = Q(x) where P(x) = 2/x. The integrating factor is μ(x) = e^(∫(2/x)dx) = e^(2ln|x|) = e^(ln(x²)) = x² (for x > 0). Multiply the entire equation by x²: x²dy/dx + 2xy = 8x³. The left side is the derivative of x²y: d/dx(x²y) = 8x³. Integrate both sides: ∫d/dx(x²y)dx = ∫8x³dx, which gives x²y = 8(x⁴/4) + C = 2x⁴ + C. Solve for y: y = (2x⁴ + C)/x² = 2x² + C/x². Apply the initial condition y(1) = 5: 5 = 2(1)² + C/(1)² = 2 + C, so C = 3. Therefore, the particular solution is y = 2x² + 3/x². The domain restriction x > 0 comes from the integrating factor calculation where we needed x ≠ 0.

Velocity is the derivative of position with respect to time, so v(t) = ds/dt. Given v(t) = 4t - 3, we have ds/dt = 4t - 3. To find position, we integrate both sides with respect to time: s(t) = ∫(4t - 3)dt. Integrating term by term: ∫4tdt = 4(t²/2) = 2t², and ∫(-3)dt = -3t. Adding the constant of integration C, we get s(t) = 2t² - 3t + C. Now we use the initial condition s(0) = 2 to find C. Substituting t = 0: s(0) = 2(0)² - 3(0) + C = C. Since s(0) = 2, we have C = 2. Therefore, the position function is s(t) = 2t² - 3t + 2. We can verify this by differentiating: ds/dt = 4t - 3, which matches the given velocity function.

We integrate both sides: ∫dy = ∫(1/x)dx, which gives y = ln|x| + C. The absolute value is necessary because the antiderivative of 1/x is defined for both positive and negative x. However, we must apply the initial condition y(1) = 0. Since x = 1 is positive, we are on the positive branch of the solution, so we can write y = ln(x) + C (without absolute value for x > 0). Applying the initial condition: 0 = ln(1) + C = 0 + C, so C = 0. Thus the particular solution is y = ln(x). The domain restriction arises because the original differential equation dy/dx = 1/x is undefined at x = 0 (division by zero). Since our initial condition is at x = 1 > 0, the solution is only guaranteed to exist on the interval (0, ∞). Therefore, the domain is x > 0. While ln|x| is defined for x

This is a separable equation. We separate variables: dy/y = 2x dx. Integrate both sides: ∫(1/y)dy = ∫2x dx, which gives ln|y| = x² + C. Exponentiating: y = e^(x² + C) = e^C · e^(x²) = Ke^(x²) where K = e^C. Apply the initial condition (0, 1): 1 = Ke^(0²) = Ke⁰= K·1, so K = 1. The particular solution is y = e^(x²). To describe its behavior: as x increases from 0, x² increases quadratically, making the exponent grow rapidly. Therefore e^(x²) grows faster than a standard exponential function; it grows at a super-exponential rate. For example, at x = 1, y = e¹ ≈ 2.718; at x = 2, y = e⁴ ≈ 54.598; at x = 3, y = e⁹ ≈ 8103. This solution exists for all real x (no domain restrictions) and increases without bound as |x| increases.

This is a linear differential equation in standard form dy/dx + P(x)y = Q(x) where P(x) = -1 and Q(x) = eˣ. First, find the integrating factor μ(x) = e^(∫P(x)dx) = e^(∫-1dx) = e⁻ˣ. Multiply the entire equation by this factor: e⁻ˣdy/dx - e⁻ˣy = e⁻ˣ·eˣ = 1. The left side is now the derivative of the product μ(x)y: d/dx(e⁻ˣy) = 1. Integrate both sides with respect to x: ∫d/dx(e⁻ˣy)dx = ∫1dx, which gives e⁻ˣy = x + C. Solve for y by multiplying both sides by eˣ: y = (x + C)eˣ = xeˣ + Ceˣ. Apply the initial condition y(0) = 1: 1 = (0 + C)e⁰= C·1, so C = 1. Substitute this value back: y = (x + 1)eˣ. We can verify this solution by differentiating: dy/dx = eˣ + (x + 1)eˣ = (x + 2)eˣ. Then dy/dx - y = (x + 2)eˣ - (x + 1)eˣ = eˣ, which matches the original equation.

This is a separable differential equation. We separate variables by multiplying both sides by (y² - 1) and dx: (y² - 1)dy = x²dx. Now integrate both sides: ∫(y² - 1)dy = ∫x²dx. The left integral becomes y³/3 - y + C₁, and the right integral becomes x³/3 + C₂. Combining the constants of integration (C = C₂ - C₁), we get y³/3 - y = x³/3 + C. Use the initial condition y(0) = 2 to find C. Substituting x = 0 and y = 2: (2³)/3 - 2 = (0³)/3 + C, which gives 8/3 - 2 = C. Converting 2 to 6/3: 8/3 - 6/3 = 2/3, so C = 2/3. The implicit solution is y³/3 - y = x³/3 + 2/3. Multiply every term by 3 to eliminate fractions: y³ - 3y = x³ + 2. This equation cannot be solved explicitly for y using elementary functions, but it correctly defines the relationship between x and y that satisfies the differential equation and initial condition.

To derive the differential equation, we analyze the rate of change of salt in the tank using dS/dt = (rate in) - (rate out). The rate of salt entering is 0 kg/min because pure water contains no salt. The rate of salt leaving equals the concentration of salt in the tank multiplied by the outflow rate. Since the volume remains constant at 1000 L (equal inflow and outflow), the concentration at time t is S(t)/1000 kg/L. With an outflow rate of 10 L/min, the rate of salt leaving is (S(t)/1000) × 10 = S(t)/100 kg/min. Therefore, dS/dt = 0 - S/100 = -S/100. This is a separable equation of the form dy/dx = g(y), describing exponential decay of salt content. We can solve it: dS/S = -dt/100, integrating gives ln|S| = -t/100 + C, so S(t) = S₀e-t/100. With initial salt S(0) = 1000 L × 0.1 kg/L = 100 kg, the solution is S(t) = 100e-t/100.