Substitution and Second-Order Enrichment

Scope Label

Enrichment / extension. Given substitutions and simple directly integrable second-order equations are useful extensions. The saved 9758 summary emphasizes first-order separable differential equations as the core route, so this note is retained for completeness and mathematical maturity rather than as the main core revision path.

Use this after First-Order Separable and Modelling.

Why This Is Separate

Both methods in this note are useful, but they do a different job from the core separable method.

• A given substitution changes the variables so that an unfamiliar first-order equation becomes familiar.
• A simple second-order equation gives information about the second derivative, so recovering $y$ requires two integrations.

Keeping these ideas in a branch prevents the hub from becoming too dense while preserving useful extension material.

Solving by a Given Substitution

Some first-order differential equations are not immediately separable, but can be reduced to a solvable form using a substitution supplied in the question.

A common substitution example is the homogeneous first-order case, where the equation depends on the ratio $\frac{y}{x}$. In that setting, the substitution $y=vx$ turns the ratio into the new variable $v$.

A common example is

$$y=vx,$$

where $v$ is treated as a function of $x$.

The purpose is reduction:

$$\text{unfamiliar DE}⟶\text{known solvable form}.$$

The Critical Derivative Step

If

$$y=vx,$$

then $v$ is not a constant. It depends on $x$.

Differentiate using the product rule:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$

This is the step students most often miss.

The substitution changes both:

• $y$ becomes $vx$;
• $\frac{dy}{dx}$ becomes $v+x\frac{dv}{dx}$.

How to read the figure. Here $v=v(x)$, so the product rule changes $dy/dx$ as well as $y$. Substitute both expressions, solve the reduced equation in $v$, then restore $y=vx$ and state any restriction such as $x\ne 0$.

In particular, for $y=vx$, the derivative relationship is

$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$

This formula is the scientific centre of the method.

Substitution Workflow

For a given substitution such as $y=vx$:

1. Differentiate the substitution with respect to $x$.
2. Replace $y$ using the substitution.
3. Replace $\frac{dy}{dx}$ using the derivative relationship.
4. Simplify the new differential equation.
5. Solve in the new variables.
6. Convert back to the original variables.

The final answer should usually be in terms of $x$ and $y$, not $v$.

Worked Example: $y=vx$

For $x\ne 0$, solve

$$\frac{dy}{dx}=1+\frac{y}{x}$$

using

$$y=vx.$$

Then

$$v=\frac{y}{x}$$

and

$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$

Since $dy/dx=v+xdv/dx$ and $y/x=v$, substitution gives

$$v+x\frac{dv}{dx}=1+v.$$

Cancelling $v$ from both sides and dividing by $x$ gives

$$\frac{dv}{dx}=\frac{1}{x}.$$

Integrating:

$$v=\ln |x|+C.$$

Therefore

$$y=x(\ln |x|+C),x\ne 0.$$

Differentiating gives $dy/dx=\ln |x|+C+1=1+y/x$, verifying the solution.

The exact algebra depends on the question, but the structural discipline is always the same: transform the derivative correctly first.

Simple Second-Order Equations

The simple second-order form treated here is

$$\frac{d^{2}y}{d{x}^2}=f(x).$$

This means the second derivative is given directly as a function of $x$.

To recover $y$, integrate twice:

$$\frac{d^{2}y}{d{x}^2}⟶\frac{dy}{dx}⟶y.$$

How to read the figure. This statement applies to the directly integrable enrichment form $y^{\prime \prime }=f(x)$, not every second-order differential equation. Two integrations introduce two independent constants, which require two independent conditions.

Why Two Constants Appear

First integration:

$$\frac{dy}{dx}=\int f(x)dx+A.$$

Second integration:

$$y=\int \left(\frac{dy}{dx}\right)dx.$$

If

$$\frac{dy}{dx}=F(x)+A,$$

then

$$y=\int [F(x)+A]dx=G(x)+Ax+B.$$

The constants $A$ and $B$ arise from the two integrations.

Worked Example: Repeated Integration

Solve

$$\frac{d^{2}y}{d{x}^2}=6x.$$

Integrate once:

$$\frac{dy}{dx}=3x^2+A.$$

Integrate again:

$$y=x^3+Ax+B.$$

This is the general solution.

If $y(0)=1$ and $y^{\prime }(1)=5$, then $B=1$ and $3+A=5$, so $A=2$. The particular solution is

$$y=x^3+2x+1.$$

It satisfies both conditions and differentiating twice returns $6x$. Conditions could instead be two values of $y$ at different points; what matters is that they are independent.

Further Enrichment: Forming a Differential Equation

An $n$-parameter family can sometimes be differentiated enough times to eliminate its arbitrary constants. For example, from $y=x^3+Ax+B$, differentiating twice gives $y^{\prime \prime }=6x$. This reverse process is useful mathematical enrichment but is not part of the core separable-equation route established by the saved syllabus summary.

Common Pitfalls

• Treating $v$ as a constant when using $y=vx$.
• Replacing $y$ but forgetting to replace $\frac{dy}{dx}$.
• Leaving the final answer in $v$ when the question asks for $x$ and $y$.
• Forgetting one of the two constants in a second-order equation.
• Applying conditions before the general solution has the correct number of constants.
• Treating enrichment methods as the core route without checking syllabus scope.

Related

• Differential Equations
• First-Order Separable and Modelling
• Integration