Solving First Order Differential Equations (Edexcel A-Level Further Mathematics): Revision Notes

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8.1.2 Solving First Order Differential Equations

General Solution of First-Order Differential Equations

A first-order differential equation has the form:

dydx=f(x,y)\frac{dy}{dx} = f(x, y)

Using substitution, we rewrite the equation in terms of new variables to simplify its form.

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Example Substitution: z=x+yz = x + y

Given a differential equation:

dydx=x+y+c1x+y+c2\frac{dy}{dx} = \frac{x + y + c_1}{x + y + c_2}

the substitution z=x+yz = x + y simplifies the equation because it eliminates x+yx + y as a combined term.

Introduction to Substitution in Differential Equations

Substitution is a powerful technique for simplifying differential equations, making them easier to solve. It is especially useful when the equation contains combinations of variables that suggest a change of variables.

Steps for Solving Using Substitution

  1. Identify an appropriate substitution z=g(x,y)z = g(x, y)
  2. Differentiate zz with respect to xx, and express dydx\frac{dy}{dx} in terms of zz and dzdx\frac{dz}{dx}
  3. Substitute into the original differential equation to rewrite it in terms of zz and dzdx\frac{dz}{dx}
  4. Solve the resulting equation for zz
  5. Back-substitute z=g(x,y)z = g(x, y) to find the solution for yy.

Finding Tangents to Polar Curves

For differential equations in polar coordinates rr and θ\theta, the slope of a tangent at any point is given by:

dydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}

Using this formula, we can find tangents that are:

Parallel to the initial line (y=0y = 0), where dydx=0\frac{dy}{dx} = 0, or:

r′sin⁡θ+rcos⁡θ=0r' \sin \theta + r \cos \theta = 0

Perpendicular to the initial line (x=0x = 0), where dydx→∞\frac{dy}{dx} \to \infty, or:

r′cos⁡θ−rsin⁡θ=0r' \cos \theta - r \sin \theta = 0

The Integrating Factor for F.O.D.E.s

An integrating factor can simplify equations of the form:

dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)

Multiplying through by the integrating factor F(x)F(x), where:

F(x)=e∫P(x) dxF(x) = e^{\int P(x) \, dx}

transforms the equation into a product rule:

F(x)dydx+F(x)P(x)y=F(x)Q(x)  ⟹  ddx[F(x)y]=F(x)Q(x)F(x) \frac{dy}{dx} + F(x) P(x) y = F(x) Q(x) \implies \frac{d}{dx}[F(x)y] = F(x)Q(x)

This allows for direct integration to solve for yy.

Worked Examples

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Example 1: Substitution in a Differential Equation

Solve dydx=x+y+3x+y−1\frac{dy}{dx} = \frac{x + y + 3}{x + y - 1} using the substitution z=x+yz = x + y

Step 1: Differentiate the Substitution

z=x+y  ⟹  dzdx=1+dydxz = x + y \implies \frac{dz}{dx} = 1 + \frac{dy}{dx}

Rearrange:

dydx=dzdx−1\frac{dy}{dx} = \frac{dz}{dx} - 1

Step 2: Substitute into the Equation

Substitute dydx=dzdx−1\frac{dy}{dx} = \frac{dz}{dx} - 1 and z=x+yz = x + y into the original equation:

dzdx−1=z+3z−1\frac{dz}{dx} - 1 = \frac{z + 3}{z - 1}

Simplify:

dzdx=z+3z−1+1=z+3+z−1z−1=2z+2z−1\frac{dz}{dx} = \frac{z + 3}{z - 1} + 1 = \frac{z + 3 + z - 1}{z - 1} = \frac{2z + 2}{z - 1}

Factorise:

dzdx=2(z+1)z−1\frac{dz}{dx} = \frac{2(z + 1)}{z - 1}

Step 3: Solve the Resulting Equation

Rearrange for integration:

∫z−1z+1 dz=∫2 dx\int \frac{z - 1}{z + 1} \, dz = \int 2 \, dx

Expand z−1z+1\frac{z - 1}{z + 1}

z−1z+1=1−2z+1\frac{z - 1}{z + 1} = 1 - \frac{2}{z + 1}

Integrate:

∫(1−2z+1) dz=∫2 dx\int (1 - \frac{2}{z + 1}) \, dz = \int 2 \, dxz−2ln⁡∣z+1∣=2x+cz - 2 \ln |z + 1| = 2x + c

Step 4: Back-Substitute z=x+yz = x + y

Substitute z=x+yz = x + y

x+y−2ln⁡∣x+y+1∣=2x+cx + y - 2 \ln |x + y + 1| = 2x + c

Simplify:

y−2ln⁡∣x+y+1∣=x+cy - 2 \ln |x + y + 1| = x + c
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Example 2: Tangents Parallel to and Perpendicular to the Initial Line

Find tangents parallel and perpendicular to the initial line for the polar curve r=1+cos⁡θr = 1 + \cos \theta

Step 1: Find r′ r′

Differentiate rr with respect to θ\theta

r′=−sin⁡θr' = -\sin \theta

Step 2: Tangents Parallel to the Initial Line (y=0y = 0)

Set:

r′sin⁡θ+rcos⁡θ=0r' \sin \theta + r \cos \theta = 0

Substitute r=1+cos⁡θr = 1 + \cos \theta

−sin⁡θ×sin⁡θ+(1+cos⁡θ)cos⁡θ=0.-\sin \theta \times \sin \theta + (1 + \cos \theta) \cos \theta = 0.

Simplify:

−sin⁡2θ+cos⁡θ+cos⁡2θ=0-\sin^2 \theta + \cos \theta + \cos^2 \theta = 0   ⟹  1−sin⁡2θ+cos⁡θ=0\implies 1 - \sin^2 \theta + \cos \theta = 0   ⟹  1+cos⁡θ=sin⁡2θ\implies 1 + \cos \theta = \sin^2 \theta

Solve for θ\theta within [0,2π][0, 2\pi]

Step 3: Tangents Perpendicular to the Initial Line (x=0x = 0)

Set:

r′cos⁡θ−rsin⁡θ=0r' \cos \theta - r \sin \theta = 0

Follow similar steps to determine θ\theta.

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Example 3: Using the Integrating Factor

Solve:

dydx+3xy=sin⁡xx3\frac{dy}{dx} + \frac{3}{x}y = \frac{\sin x}{x^3}

Step 1: Identify P(x)P(x)

Here, P(x)=3xP(x) = \frac{3}{x}

Step 2: Find the Integrating Factor

F(x)=e∫P(x) dx=e∫3x dx=e3ln⁡x=x3F(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln x} = x^3

Step 3: Multiply Through by F(x)F(x)

x3dydx+3x2y=sin⁡xx^3 \frac{dy}{dx} + 3x^2 y = \sin x

Rewrite as:

ddx(x3y)=sin⁡x\frac{d}{dx}(x^3 y) = \sin x

Step 4: Integrate

x3y=∫sin⁡x dx=−cos⁡x+cx^3 y = \int \sin x \, dx = -\cos x + c

Solve for yy:

y=−cos⁡x+cx3y = \frac{-\cos x + c}{x^3}

Note Summary

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Common Mistakes

  1. Forgetting to differentiate substitutions correctly (e.g.,z=x+y z = x + y)
  2. Skipping steps when rearranging differential equations.
  3. Neglecting to back-substitute the original variables after solving.
  4. Misapplying the integrating factor F(x)F(x) by omitting e∫P(x) dxe^{\int P(x) \, dx}
  5. Incorrectly evaluating boundary conditions or constants of integration.
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Key Formulas

  1. Substitution Rule:
z=g(x,y)  ⟹  dzdx=∂z∂x+∂z∂ydydxz = g(x, y) \quad \implies \quad \frac{dz}{dx} = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{dy}{dx}
  1. Area in Polar Coordinates:
A=12∫αβr2 dθA = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta
  1. Slope of Tangent in Polar Coordinates:
dydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}
  1. Integrating Factor:
F(x)=e∫P(x) dxF(x) = e^{\int P(x) \, dx}

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