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Calculus - Differential equations (HL)

Grade 11IB_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.

Differential equations (HL) Grade 11 Notes & Examples