What does it mean to reduce a first-order differential equation?

Reducing an equation means using a given substitution to turn a hard first-order equation into an easier one, written in new variables. The point is that the easier equation is one a standard method can handle, such as direct integration, separation of variables or an integrating factor, while the original is none of those.

True or False? Reducing a second-order differential equation leaves it second order.

True. Reducing here means making the equation easier to solve rather than lowering its order, so a second-order equation transforms into another second-order equation. What the transformation removes is the difficulty, not a derivative.

A LevelFurther MathsEdexcelFurther Pure 1FlashcardsFurther Differential EquationsReducing Differential Equations

Reducing Differential Equations (Edexcel A Level Further Maths: Further Pure 1): Flashcards

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FrontReducing First-Order Differential Equations
What does it mean to reduce a first-order differential equation?

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What does it mean to reduce a first-order differential equation?
Reducing an equation means using a given substitution to turn a hard first-order equation into an easier one, written in new variables.
The point is that the easier equation is one a standard method can handle, such as direct integration, separation of variables or an integrating factor, while the original is none of those.
In $\frac{d x}{d t} = x + t^{2}$ , which variable is dependent and which is independent?
Here $x$ is the dependent variable and $t$ is the independent one, because the solution will take the form $x = g (t)$ .
Read it off the derivative rather than off the letters: whatever sits on top of the derivative is the dependent variable, and in a modelling question that is often not $y$ .
True or False?
Transforming the independent variable leaves the dependent variable unchanged.
True.
A transformation of the independent variable replaces only the variable that is being depended on, so the dependent variable carries straight through the working untouched.
For example, changing $(x, y)$ into $(t, y)$ leaves $y$ alone, and the final answer is still a formula for $y$ .
A transformation of the dependent variable changes $(x, y)$ into $(x, z)$ . Complete the chain rule that rewrites the derivative:
$\frac{d y}{d x} =______\times \frac{d z}{d x}$
The completed chain rule is:
$\frac{d y}{d x} = \frac{d y}{d z} \times \frac{d z}{d x}$
The missing factor is the one the transformation itself supplies: differentiating $y = h (z)$ gives $\frac{d y}{d z}$ straight away.
When transforming the independent variable using $x = \frac{1}{t}$ , why is it easier to find $\frac{d t}{d x}$ from $\frac{d x}{d t}$ than directly?
Because the transformation already gives $x$ in terms of $t$ , so $\frac{d x}{d t}$ can be written down at once and then inverted using $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$ .
Here $\frac{d x}{d t} = - t^{- 2}$ , so $\frac{d t}{d x} = - t^{2}$ , whereas differentiating directly would mean rearranging the transformation first.
A differential equation is to be reduced using the product transformation $y = x z$ . Which rule gives $\frac{d y}{d x}$ , and what is the result?
The product rule gives $\frac{d y}{d x} = z + x \frac{d z}{d x}$ , since $z$ is itself a function of $x$ .
A quotient such as $y = \frac{z}{x}$ is handled the same way with the quotient rule, and a transformation such as $y = x^{2} z^{3}$ needs implicit differentiation on top of it.
True or False?
When a substitution is used to reduce a differential equation, replacing the derivative is enough, and any $y$ terms still left in the equation can stay.
False.
The transformed equation has to be written in the new variables only, so every remaining $y$ must be converted using the transformation itself before anything can be solved.
In $\frac{d y}{d x} = x y \ln y$ with $y = e^{z}$ , both the $y$ and the $\ln y$ on the right still have to go.
A differential equation has been reduced and the easier equation solved, giving $z$ in terms of $x$ . What are the last two steps, and in which order?
Use the transformation to convert that solution back into $y$ and $x$ , and only then substitute the boundary conditions.
The conditions are values of the original variables, so using them any earlier would mean putting an $x$ and $y$ pair into an equation written in $z$ .
True or False?
Reducing a second-order differential equation leaves it second order.
True.
Reducing here means making the equation easier to solve rather than lowering its order, so a second-order equation transforms into another second-order equation.
What the transformation removes is the difficulty, not a derivative.
What form is a second-order differential equation being reduced to?
The target is $a \frac{d^{2} z}{d x^{2}} + b \frac{d z}{d x} + c z = f (x)$ , in which $a$ , $b$ and $c$ are constants.
Only a constant-coefficient equation has an auxiliary equation to solve, so a transformation earns its place exactly when it clears variable coefficients such as $x^{2}$ and $5 x$ out of the way.
In reducing a second-order equation you have already written $\frac{d y}{d x}$ in terms of the new variable. How do you now obtain $\frac{d^{2} y}{d x^{2}}$ ?
Differentiate both sides of that result with respect to $x$ again, since $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x})$ .
The right-hand side is a product of two things that both change with $x$ , so the product rule is needed at this step almost every time.
Part-way through reducing a second-order equation, a factor written in $z$ has to be differentiated with respect to $x$ . Complete the identity that makes that possible:
$\frac{d}{d x} (z^{- 2}) = \frac{d}{d z} (z^{- 2}) \times______$
The completed identity is:
$\frac{d}{d x} (z^{- 2}) = \frac{d}{d z} (z^{- 2}) \times \frac{d z}{d x}$
Anything written in $z$ can only be differentiated with respect to $z$ , and the extra factor $\frac{d z}{d x}$ is what converts that into a derivative with respect to $x$ .
The transformation $y = \frac{1}{z}$ gives $\frac{d y}{d x} = - \frac{1}{z^{2}} \frac{d z}{d x}$ . Why does differentiating this again produce a term in ${(\frac{d z}{d x})}^{2}$ ?
Because the product rule differentiates the factor $- \frac{1}{z^{2}}$ , and that factor is written in $z$ , so doing so picks up a further $\frac{d z}{d x}$ .
That new factor then multiplies the $\frac{d z}{d x}$ already standing beside it, giving the term $\frac{2}{z^{3}} {(\frac{d z}{d x})}^{2}$ .
Why is $x = e^{t}$ the right transformation for $x^{2} \frac{d^{2} y}{d x^{2}} + 5 x \frac{d y}{d x} + 4 y = \ln x$ ?
Because each coefficient is the power of $x$ matching the order of its derivative, and $x = e^{t}$ makes each transformed derivative carry an $e^{- t}$ or $e^{- 2 t}$ factor that those powers cancel exactly.
The equation becomes $\frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 4 y = t$ , with the right-hand side simplifying too, since $\ln (e^{t}) = t$ .
The general solution has been transformed back to $y$ in terms of $x$ . What extra work does a condition on $\frac{d y}{d x}$ need that a condition on $y$ does not?
The general solution has to be differentiated with respect to $x$ before the condition can be substituted, which a condition on $y$ alone never requires.
That differentiation is done on the transformed-back solution, so the product rule often reappears on terms such as $B x^{- 2} \ln x$ .

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