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Calculus - Differential equations (HL)

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A first-order differential equation is an equation of the form dydx=f(x,y)\frac{dy}{dx} = f(x, y). The general solution contains an arbitrary constant CC.

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Separable differential equations can be written in the form dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y). These are solved by integrating both sides: ∫1h(y)dy=∫g(x)dx\int \frac{1}{h(y)} dy = \int g(x) dx.

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Homogeneous differential equations of the form dydx=f(yx)\frac{dy}{dx} = f\left(\frac{y}{x}\right) can be solved using the substitution y=vxy = vx, which implies dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}.

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First-order linear differential equations have the form dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x). These are solved using an integrating factor I(x)=e∫P(x)dxI(x) = e^{\int P(x) dx}.

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Euler's method is a numerical technique to approximate solutions to dydx=f(x,y)\frac{dy}{dx} = f(x, y) given an initial point (x0,y0)(x_0, y_0) and a step size hh.

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The Maclaurin series method can be used to find power series solutions to differential equations by repeatedly differentiating the original equation to find higher-order derivatives at x=0x = 0.

📐Formulae

∫1g(y)dy=∫f(x)dx+C\int \frac{1}{g(y)} dy = \int f(x) dx + C

I(x)=e∫P(x)dxI(x) = e^{\int P(x) dx}

y⋅I(x)=∫Q(x)⋅I(x)dxy \cdot I(x) = \int Q(x) \cdot I(x) dx

yn+1=yn+h×f(xn,yn)y_{n+1} = y_n + h \times f(x_n, y_n)

v+xdvdx=f(v) where y=vxv + x\frac{dv}{dx} = f(v) \text{ where } y = vx

💡Examples

Problem 1:

Solve the differential equation dydx=x2y\frac{dy}{dx} = \frac{x^2}{y} given that y(0)=2y(0) = 2.

Solution:

∫ydy=∫x2dx\int y dy = \int x^2 dx y22=x33+C\frac{y^2}{2} = \frac{x^3}{3} + C Substitute x=0,y=2x=0, y=2: 222=0+C  ⟹  C=2\frac{2^2}{2} = 0 + C \implies C = 2 y22=x33+2\frac{y^2}{2} = \frac{x^3}{3} + 2 y2=2x33+4  ⟹  y=2x33+4y^2 = \frac{2x^3}{3} + 4 \implies y = \sqrt{\frac{2x^3}{3} + 4}

Explanation:

This is a separable differential equation. We group all yy terms with dydy and xx terms with dxdx, integrate both sides, and use the initial condition to find the particular constant CC.

Problem 2:

Find the general solution of the linear differential equation dydx+2xy=4x\frac{dy}{dx} + \frac{2}{x}y = 4x.

Solution:

Identify P(x)=2xP(x) = \frac{2}{x}. Calculate the integrating factor: I(x)=e∫2xdx=e2ln⁡∣x∣=x2I(x) = e^{\int \frac{2}{x} dx} = e^{2\ln|x|} = x^2 Multiply the DE by I(x)I(x): x2dydx+2xy=4x3x^2 \frac{dy}{dx} + 2xy = 4x^3 ddx(x2y)=4x3\frac{d}{dx}(x^2 y) = 4x^3 Integrate both sides: x2y=∫4x3dx=x4+Cx^2 y = \int 4x^3 dx = x^4 + C Divide by x2x^2: y=x2+Cx2y = x^2 + \frac{C}{x^2}

Explanation:

This is a first-order linear differential equation. We use the Integrating Factor method to convert the left side into the derivative of a product (I(x)⋅y)(I(x) \cdot y).

Problem 3:

Use Euler's method with a step size of h=0.1h = 0.1 to approximate y(0.2)y(0.2) for the differential equation dydx=x+y\frac{dy}{dx} = x + y with y(0)=1y(0) = 1.

Solution:

Step 1: x0=0,y0=1,f(x,y)=x+yx_0 = 0, y_0 = 1, f(x,y) = x + y y1=y0+h(x0+y0)=1+0.1(0+1)=1.1y_1 = y_0 + h(x_0 + y_0) = 1 + 0.1(0 + 1) = 1.1 Step 2: x1=0.1,y1=1.1x_1 = 0.1, y_1 = 1.1 y2=y1+h(x1+y1)=1.1+0.1(0.1+1.1)=1.1+0.1(1.2)=1.22y_2 = y_1 + h(x_1 + y_1) = 1.1 + 0.1(0.1 + 1.1) = 1.1 + 0.1(1.2) = 1.22 Therefore, y(0.2)≈1.22y(0.2) \approx 1.22.

Explanation:

Euler's method is applied iteratively. Each new yy value is calculated by adding the product of the step size and the gradient at the current point to the previous yy value.