First-Order Linear Equations

What is a linear first order equation?

A linear first order equation is an equation in the form

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}+P(x)y=Q(x)} .

The simpliest case of which is shown below in Example 1 where $P(x)$ and $Q(x)$ are not functions but simple constants. Linear first order equations are important because they show up frequently in nature and physics, and can be solved by a fairly straight forward method.

We first begin by motivating the method. First recall that the product rule states that $[f(x)\cdot g(x)]'=f'(x)g(x)+f(x)g'(x)$. The key observation is that the left hand side of the first order ODE appears to be fairly similar to the product rule. There is one term with a derivative of $y$, and there is one term without a derivative of $y$.

Let's compare the left hand side

${\frac {dy}{dx}}+P(x)y$

to the product rule applied to the product involving Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} . Notice that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}} is equal to the prime Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle '} operator.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}\Big[\mu(x)y(x)\Big]=\mu(x)\frac{dy}{dx}+\frac{d\mu}{dx}y(x)} .

The first fact we notice is that on the left hand side of our first order ODE there is no Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)} in front of the term Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}} . We can try to correct that by multiplying through by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)} . Then the ODE would become:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)\frac{dy}{dx}+\mu(x)P(x)y=\mu(x)Q(x).}

But now this would only look like the product rule if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\mu}{dx}=\mu(x)P(x)} and hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)\frac{dy}{dx}+y(x)\frac{d\mu}{dx} = \mu(x)Q(x)} , which we do not know by now. But this simple equation can easily be solved, as it is a separable equation. Dividing both sides of it by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)} and integrating we get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{1}{\mu(x)}\frac{d\mu}{dx}\,dx = \int P(x)\,dx}

Integrating by substitution the integral on the left hand side is simply Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln |\mu(x)|} . Thus pulling over the natural logarithm to the right, we have:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)=e^{\int P(x)\,dx}.}

This is very important and is called the 'integrating factor'. So our original ODE can be rewritten as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)\,dx}\frac{dy}{dx}+P(x)e^{\int P(x)\,dx }y=e^{\int P(x)\,dx}Q(x)}

But now the whole point of introducing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu(x)} was we were trying to turn the left hand side of this ODE into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\mu(x)y]'} for easily integrating it, which is exactly what we have done. Thus

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [e^{\int P(x)\,dx}y]'=e^{\int P(x)\,dx}Q(x).}

So we may integrate both sides and then solve for y to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)\,dx}y=\int e^{\int P(x)\,dx}Q(x)\,dx.}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=e^{-\int P(x)\,dx}\int e^{\int P(x)\,dx}Q(x)\,dx.}

Below we provide another derivation of this formula.

Example 1: P and Q are Constant

Let's say we have the equation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}= my + n}

where n and m are constants. Solve this for y.

Step 1: Find the integrating factor, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}} ,

with P(x) = m and the integration of a constant being

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int mdx=mx+C}

yielding

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}=e^{mx}e^c=Ce^{mx}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^c} is collapsing into C. Letting C=1, we get e^mx.

Step 2: Multiply through by the integrating factor just found.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{mx}\frac{dy}{dx}+e^{mx}my=ne^{mx} \,}

Step 3: Recognize that the left hand side is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \Big[e^{\int P(x)dx}y\Big]} ,

that is subject to the product rule, giving us.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \Big[e^{mx}y\Big]=ne^{mx}}

Step 4: Integrate both sides with respect to x,

with the left hand side now easily integrated.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int (\frac{d}{dx} e^{mx}y)dx=\int ne^{mx}dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{mx}y=\frac{n}{m}e^{mx}+C}

Step 5: And finally solve for y

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=\frac{n}{m}+Ce^{-mx}}

Example 2: P and Q are x

Take the equation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}+xy=x}

Solve for y.

Step 1: Find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}} , P(x) = x and the integral of x being

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x=\frac{1}{2}x^2+C}

yielding

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}=Ce^{\frac{1}{2}x^2}}

Letting C=1, we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\frac{1}{2}x^2}}

Step 2: Multiply through

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\frac{1}{2}x^2}\frac{dy}{dx}+e^{\frac{1}{2}x^2}xy=e^{\frac{1}{2}x^2}x}

Step 3: Recognize that the left hand is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} e^{\int P(x)dx}y} , giving

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} e^{\frac{1}{2}x^2}y=e^{\frac{1}{2}x^2}x}

Step 4: Integrate

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int (\frac{d}{dx} e^{\frac{1}{2}x^2}y)dx=\int xe^{\frac{1}{2}x^2}dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\frac{1}{2}x^2}y=e^{\frac{1}{2}x^2}+C}

Step 5: Solve for y

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=1+\frac{C}{e^{\frac{1}{2}x^2}}}

Example 3: P and Q are Unrelated

Take the equation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}+\frac{1}{x}y=x}

Solve for y.

Before we begin we notice that for the function y, we are not guaranteed a solution for all x because the coefficient is discontinuous at 0. So either we can find a solution in $x\in (0,\infty )$ or for $x\in (-\infty ,0)$. For the purposes of this solution we shall assume Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in (0,\infty)} .

Step 1: Find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x) = 1/x} ,

and the integration of 1/x being

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{x}\,dx=\ln(x)+C}

yielding

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}=Ce^{\ln(x)}=Cx}

Letting C=1, we get x.

Step 2: Multiply through

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\frac{dy}{dx}+y=x^2}

Step 3: Recognize that the left hand is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} e^{\int P(x)dx}y}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} [xy]=x^2}

Step 4: Integrate both sides w.r.t.x.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{d}{dx} [xy]\,dx=\int x^2dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xy=\frac{1}{3}x^3+C}

Step 5: Solve for y

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=\frac{1}{3}x^2+\frac{C}{x}}

An alternative derivation

There is a strategy for solving inhomogeneous equations called variation of parameters that proceeds as follows. First find the general solution to the homogeneous equations. Guess that the solutions of the inhomogeneous equations can be written in the same way as for the solutions of the homogeneous equations, except where the unknown constants are replaced by unknown functions. Then you plug a solution of this form back in to the equation to see if you can find what the unknown function is.

This method works well in case of first order linear equations and gives us an alternative derivation of our formula for the solution which we present below.

1. First, set Q(x) equal to 0 so that you end up with a homogeneous linear equation (the usage of this term is to be distinguished from the usage of "homogeneous" in the previous sections).
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}+P(x)y=0}
   
   
2. This equation is separable, so separate it:
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{y}+P(x)dx=0}
   
   
3. Solve the equation to obtain the solution
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=Ce^{-\int P(x) dx}}
   
   
4. Now let C be replaced by a variable function of x, and denote it g(x).
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=g(x)e^{-\int P(x) dx}}
   
   
5. Substitute the previous equation into the differential equation to get
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)e^{-\int P(x) dx}=Q(x)}
   
   
6. Now solve for g(x) to get
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=C+\int Q(x)e^{\int P(x) dx}dx}
   
   

1. Now obtain the general solution by plugging in this expression in g(x):
   
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=g(x)e^{-\int P(x) dx}=Ce^{-\int P(x) dx}+e^{-\int P(x) dx}\int Q(x)e^{\int P(x) dx}dx}

Making Linear Equations from Non-Linear Equations

Sometimes a non-linear equation, which is not solvable like this, can be made linear, and more easily solvable, by applying a substitution.

Example

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y\frac{dy}{dx}+xy^2=5x}

Let's make the following substitution:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=y^2 \,}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dv}{dx}=2y\frac{dy}{dx}}

Plugging in, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}\frac{dv}{dx}+xv=5x}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dv}{dx}+2xv=10x}

We can then solve as a linear equation in v, using the step-by-step method above:

Step 1: Find the integrating factor:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int 2x dx}=e^{x^2+C}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\int P(x)dx}=Ce^{x^2}}

Letting C=1 for convenience, we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{x^2}} as our integrating factor.

Step 2: Multiply through

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{x^2}\frac{dv}{dx}+e^{x^2}xv=10e^{x^2}x}

Step 3: Recognize that the left hand is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} e^{\int P(x)dx}v}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} e^{x^2}v=10e^{x^2}x}

Step 4: Integrate both sides w.r.t.x.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int (\frac{d}{dx} e^{x^2}v)dx=\int 10e^{x^2}xdx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{x^2}v=5e^{x^2}+C}

Step 5: Solve for v.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=5+\frac{C}{e^{x^2}}}

Now that we have v, solve for y.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=y^2 \,}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y^2=5+\frac{C}{e^{x^2}} \,}

Resources

• Solving Linear First-Order Differential Equations Video by James Sousa, Math is Power 4U
• Ex 1: Solve a Linear First-Order Differential Equation Video by James Sousa, Math is Power 4U
• Ex 2: Solve a Linear First-Order Differential Equation Video by James Sousa, Math is Power 4U
• Ex 3: Solve a Linear First-Order Differential Equation Video by James Sousa, Math is Power 4U
• Ex 1: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U
• Ex 2: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U
• Ex 3: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U
• Ex 4: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U
• Ex 5: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U
• Ex 6: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U
• Ex 7: Solve a Linear First-Order Differential Equation Using an Integrating Factor Video by James Sousa, Math is Power 4U

• Integrating Factors 1 Video by Khan Academy
• Integrating Factors 2 Video by Khan Academy

• First Order Linear Differential Equations Video by PatrickJMT
• Integrating Factor to Solve a First Order Linear Differential Equation Video by PatrickJMT

• First Order Linear Differential Equations Video by The Organic Chemistry Tutor

Licensing

Content obtained and/or adapted from:

• [] under a CC BY-SA license