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First-Order Linear Equations
Write a first-order linear differential equation in standard form.
Standard Form
Consider the differential equation
\[(3{x}^{2}-4){y}^{'}+(x-3)y=\text{sin}\ x.\]Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of \({y}^{'}\) be equal to \(1.\) To make this happen, we divide both sides by \(3{x}^{2}-4.\)
\[{y}^{'}+(\frac{x-3}{3{x}^{2}-4})\ y=\frac{\text{sin}\ x}{3{x}^{2}-4}\]This is called the standard form of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to , we can divide both sides of the equation by \(a(x).\) This leads to the equation
\[{y}^{'}+\frac{b(x)}{a(x)}y=\frac{c(x)}{a(x)}.\]Now define \(p(x)=\frac{b(x)}{a(x)}\) and \(q(x)=\frac{c(x)}{a(x)}.\) Then becomes
\[{y}^{'}+p(x)y=q(x).\]We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.
Example
Try it.
Put each of the following first-order linear differential equations into standard form. Identify \(p(x)\) and \(q(x)\) for each equation.
- \(y'=3x-4y\)
- \(\frac{3xy'}{4y-3}=2\) (here \(x\ne 0)\)
- \(y=3y'-4{x}^{2}+5\)
Solution
- Add \(4y\) to both sides:
\[y'+4y=3x.\]
In this equation, \(p(x)=4\) and \(q(x)=3x.\) - Multiply both sides by \(4y-3,\) then subtract \(8y\) from each side:
\[\begin{array}{lll}\frac{3xy'}{4y-3} & = & 2 \\ 3xy' & = & 2(4y-3) \\ 3xy' & = & 8y-6 \\ 3xy'-8y & = & -6.\end{array}\]
Finally, divide both sides by \(3x\) to make the coefficient of \(y'\) equal to \(1\text{:}\)
\[y'-\frac{8}{3x}y=-\frac{2}{x}.\] This is allowable because in the original statement of this problem we assumed that \(x\ne 0.\) (If \(x=0\) then the original equation becomes \(0=2,\) which is clearly a false statement.)
In this equation, \(p(x)=-\frac{8}{3x}\) and \(q(x)=-\frac{2}{x}.\) - Subtract \(y\) from each side and add \(4{x}^{2}-5\text{:}\)
\[3y'-y=4{x}^{2}-5.\]
Next divide both sides by \(3\text{:}\)
\[y'-\frac{1}{3}y=\frac{4}{3}{x}^{2}-\frac{5}{3}.\]
In this equation, \(p(x)=-\frac{1}{3}\) and \(q(x)=\frac{4}{3}{x}^{2}-\frac{5}{3}.\)
Integrating Factors
We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:
\[y'+p(x)y=q(x).\]The first term on the left-hand side of is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the product rule from the Differentiation Rules section. If we multiply by a yet-to-be-determined function \(\mu (x),\) then the equation becomes
\[\mu (x){y}^{'}+\mu (x)p(x)y=\mu (x)q(x).\]The left-hand side can be matched perfectly to the product rule:
\[\frac{d}{dx}[f(x)g(x)]={f}^{'}(x)g(x)+f(x){g}^{'}(x).\]Matching term by term gives \(y=f(x),g(x)=\mu (x),\) and \({g}^{'}(x)=\mu (x)p(x).\) Taking the derivative of \(g(x)=\mu (x)\) and setting it equal to the right-hand side of \({g}^{'}(x)=\mu (x)p(x)\) leads to
\[{\mu }^{'}(x)=\mu (x)p(x).\]This is a first-order, separable differential equation for \(\mu (x).\) We know \(p(x)\) because it appears in the differential equation we are solving. Separating variables and integrating yields
\[\begin{array}{lll}\frac{{\mu }^{'}(x)}{\mu (x)} & = & p(x) \\ \int \frac{{\mu }^{'}(x)}{\mu (x)}dx & = & \int p(x)\ dx \\ \text{ln}|\mu (x)| & = & \int p(x)\ dx+C \\ {e}^{\text{ln}|\mu (x)|} & = & {e}^{\int p(x)\ dx+C} \\ |\mu (x)| & = & {C}_{1}{e}^{\int p(x)\ dx} \\ \mu (x) & = & {C}_{2}{e}^{\int p(x)\ dx}.\end{array}\]Here \({C}_{2}\) can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of by the integrating factor \(\mu (x).\) This gives
\[\mu (x){y}^{'}+\mu (x)p(x)y=\mu (x)q(x).\]The left-hand side of can be rewritten as \(\frac{d}{dx}(\mu (x)y).\)
\[\frac{d}{dx}(\mu (x)y)=\mu (x)q(x).\]\[\begin{array}{lll}\int \frac{d}{dx}(\mu (x)y)\ dx & = & \int \mu (x)q(x)\ dx \\ \mu (x)y & = & \int \mu (x)q(x)\ dx.\end{array}\]\[y=\frac{1}{\mu (x)}[\int \mu (x)q(x)\ dx+C].\]Condensed — the full section is in OpenStax Calculus Volume 2.
Applications of First-order Linear Differential Equations
We look at two different applications of first-order linear differential equations. The first involves air resistance as it relates to objects that are rising or falling; the second involves an electrical circuit. Other applications are numerous, but most are solved in a similar fashion.
Condensed — the full section is in OpenStax Calculus Volume 2.
Key Concepts
- Any first-order linear differential equation can be written in the form \(y'+p(x)y=q(x).\)
- We can use a five-step problem-solving strategy for solving a first-order linear differential equation that may or may not include an initial value.
- Applications of first-order linear differential equations include determining motion of a rising or falling object with air resistance and finding current in an electrical circuit.
First-Order Linear Equations
Are the following differential equations linear? Explain your reasoning.
Write the following first-order differential equations in standard form.
What are the integrating factors for the following differential equations?
Solve the following differential equations by using integrating factors.
Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?
Solve the following initial-value problems by using integrating factors.
For the following problems, determine how parameter \(a\) affects the solution.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Put each of the following first-order linear differential equations into standard form. Identify \(p(x)\) and \(q(x)\) for each equation.
- \(y'=3x-4y\)
- \(\frac{3xy'}{4y-3}=2\) (here \(x\ne 0)\)
- \(y=3y'-4{x}^{2}+5\)
जवाब दिखाएँ
- Add \(4y\) to both sides:
\[y'+4y=3x.\]
In this equation, \(p(x)=4\) and \(q(x)=3x.\) - Multiply both sides by \(4y-3,\) then subtract \(8y\) from each side:
\[\begin{array}{lll}\frac{3xy'}{4y-3} & = & 2 \\ 3xy' & = & 2(4y-3) \\ 3xy' & = & 8y-6 \\ 3xy'-8y & = & -6.\end{array}\]
Finally, divide both sides by \(3x\) to make the coefficient of \(y'\) equal to \(1\text{:}\)
\[y'-\frac{8}{3x}y=-\frac{2}{x}.\] This is allowable because in the original statement of this problem we assumed that \(x\ne 0.\) (If \(x=0\) then the original equation becomes \(0=2,\) which is clearly a false statement.)
In this equation, \(p(x)=-\frac{8}{3x}\) and \(q(x)=-\frac{2}{x}.\) - Subtract \(y\) from each side and add \(4{x}^{2}-5\text{:}\)
\[3y'-y=4{x}^{2}-5.\]
Next divide both sides by \(3\text{:}\)
\[y'-\frac{1}{3}y=\frac{4}{3}{x}^{2}-\frac{5}{3}.\]
In this equation, \(p(x)=-\frac{1}{3}\) and \(q(x)=\frac{4}{3}{x}^{2}-\frac{5}{3}.\)
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Put the equation \(\frac{(x+3)y'}{2x-3y-4}=5\) into standard form and identify \(p(x)\) and \(q(x).\)
जवाब दिखाएँ
\(y'+\frac{15}{x+3}y=\frac{10x-20}{x+3};p(x)=\frac{15}{x+3}\) and \(q(x)=\frac{10x-20}{x+3}\)
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Find a general solution for the differential equation \(xy'+3y=4{x}^{2}-3x.\) Assume \(x>0.\)
जवाब दिखाएँ
- To put this differential equation into standard form, divide both sides by \(x\text{:}\)
\[y'+\frac{3}{x}y=4x-3.\]
Therefore \(p(x)=\frac{3}{x}\) and \(q(x)=4x-3.\) - The integrating factor is \(\mu (x)={e}^{\int (3\text{/}x)\ dx}={e}^{3\ \text{ln}\ x}={x}^{3}.\)
- Multiplying both sides of the differential equation by \(\mu (x)\) gives us
\[\begin{array}{lll}{x}^{3}{y}^{'}+{x}^{3}(\frac{3}{x})y & = & {x}^{3}(4x-3) \\ {x}^{3}{y}^{'}+3{x}^{2}y & = & 4{x}^{4}-3{x}^{3} \\ \frac{d}{dx}({x}^{3}y) & = & 4{x}^{4}-3{x}^{3}.\end{array}\] - Integrate both sides of the equation.
\[\begin{array}{lll}\int \frac{d}{dx}({x}^{3}y)dx & = & \int 4{x}^{4}-3{x}^{3}dx \\ {x}^{3}y & = & \frac{4{x}^{5}}{5}-\frac{3{x}^{4}}{4}+C \\ y & = & \frac{4{x}^{2}}{5}-\frac{3x}{4}+C{x}^{-3}.\end{array}\] - There is no initial value, so the problem is complete.
- To put this differential equation into standard form, divide both sides by \(x\text{:}\)
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Find the general solution to the differential equation \((x-2)y'+y=3{x}^{2}+2x.\) Assume \(x>2.\)
जवाब दिखाएँ
\(y=\frac{{x}^{3}+{x}^{2}+C}{x-2}\)
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Solve the initial-value problem
\[{y}^{'}+3y=2x-1,\ y(0)=3.\]जवाब दिखाएँ
- This differential equation is already in standard form with \(p(x)=3\) and \(q(x)=2x-1.\)
- The integrating factor is \(\mu (x)={e}^{\int 3dx}={e}^{3x}.\)
- Multiplying both sides of the differential equation by \(\mu (x)\) gives
\[\begin{array}{lll}{e}^{3x}{y}^{'}+3{e}^{3x}y & = & (2x-1){e}^{3x} \\ \frac{d}{dx}[y{e}^{3x}] & = & (2x-1){e}^{3x}.\end{array}\]
Integrate both sides of the equation:
\[\begin{array}{lll}\int \frac{d}{dx}[y{e}^{3x}]dx & = & \int (2x-1){e}^{3x}dx \\ y{e}^{3x} & = & \frac{{e}^{3x}}{3}(2x-1)-\int \frac{2}{3}{e}^{3x}dx \\ y{e}^{3x} & = & \frac{{e}^{3x}(2x-1)}{3}-\frac{2{e}^{3x}}{9}+C \\ y & = & \frac{2x-1}{3}-\frac{2}{9}+C{e}^{-3x} \\ y & = & \frac{2x}{3}-\frac{5}{9}+C{e}^{-3x}.\end{array}\] - Now substitute \(x=0\) and \(y=3\) into the general solution and solve for \(C\text{:}\)
\[\begin{array}{lll}y & = & \frac{2}{3}x-\frac{5}{9}+C{e}^{-3x} \\ 3 & = & \frac{2}{3}(0)-\frac{5}{9}+C{e}^{-3(0)} \\ 3 & = & -\frac{5}{9}+C \\ C & = & \frac{32}{9}.\end{array}\]
Therefore the solution to the initial-value problem is
\[y=\frac{2}{3}x-\frac{5}{9}+\frac{32}{9}{e}^{-3x}.\]
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Solve the initial-value problem \(y'-2y=4x+3\ y(0)=-2.\)
जवाब दिखाएँ
\(y=-2x-\frac{5}{2}+\frac{1}{2}{e}^{2x}\)
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A racquetball is hit straight upward with an initial velocity of \(2\) m/s. The mass of a racquetball is approximately \(0.0427\) kg. Air resistance acts on the ball with a force numerically equal to \(0.5v,\) where \(v\) represents the velocity of the ball at time \(t.\)
- Find the velocity of the ball as a function of time.
- How long does it take for the ball to reach its maximum height?
- If the ball is hit from an initial height of \(1\) meter, how high will it reach?
जवाब दिखाएँ
- The mass \(m=0.0427\ \text{kg},k=0.5,\) and \(g=9.8{\ \text{m/s}}^{2}.\) The initial velocity is \({v}_{0}=2\) m/s. Therefore the initial-value problem is
\[0.0427\frac{dv}{dt}=-0.5v-0.0427(9.8),\ {v}_{0}=2.\]
Dividing the differential equation by \(0.0427\) gives
\[\frac{dv}{dt}=-11.7096v-9.8,\ {v}_{0}=2.\]
The differential equation is linear. Using the problem-solving strategy for linear differential equations:
Step 1. Rewrite the differential equation as \(\frac{dv}{dt}+11.7096v=-9.8.\) This gives \(p(t)=11.7096\) and \(q(t)=-9.8\)
Step 2. The integrating factor is \(\mu (t)={e}^{\int 11.7096dt}={e}^{11.7096t}.\)
Step 3. Multiply the differential equation by \(\mu (t)\text{:}\)
\[\begin{array}{lll}{e}^{11.7096t}\frac{dv}{dt}+11.7096v{e}^{11.7096t} & = & -9.8{e}^{11.7096t} \\ \frac{d}{dt}[v{e}^{11.7096t}] & = & -9.8{e}^{11.7096t}.\end{array}\]
Step 4. Integrate both sides:
\[\begin{array}{lll}\int \frac{d}{dt}[v{e}^{11.7096t}]\ dt & = & \int -9.8{e}^{11.7096t}dt \\ v{e}^{11.7096t} & = & \frac{-9.8}{11.7096}{e}^{11.7096t}+C \\ v(t) & = & -0.8369+C{e}^{-11.7096t}.\end{array}\]
Step 5. Solve for \(C\) using the initial condition \({v}_{0}=v(0)=2\text{:}\)
\[\begin{array}{lll}v(t) & = & -0.8369+C{e}^{-11.7096t} \\ v(0) & = & -0.8369+C{e}^{-11.7096(0)} \\ 2 & = & -0.8369+C \\ C & = & 2.8369.\end{array}\]
Therefore the solution to the initial-value problem is \(v(t)=2.8369{e}^{-11.7096t}-0.8369.\) - The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:
\[\begin{array}{lll} \\ 2.8369{e}^{-11.7096t}-0.8369 & = & 0 \\ 2.8369{e}^{-11.7096t} & = & 0.8369 \\ {e}^{-11.7096t} & = & \frac{0.8369}{2.8369}\approx 0.295 \\ \text{ln}\ {e}^{-11.7096t} & = & \text{ln}\ 0.295\approx -1.221 \\ -11.7096t & = & -1.221 \\ t & \approx & 0.104.\end{array}\]
Therefore it takes approximately \(0.104\) second to reach maximum height. - To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if \(h(t)\) represents the height at time \(t,\) then \({h}^{'}(t)=v(t).\) Because we know \(v(t)\) and the initial height, we can form an initial-value problem:
\[{h}^{'}(t)=2.8369{e}^{-11.7096t}-0.8369,\ h(0)=1.\]
Integrating both sides of the differential equation with respect to \(t\) gives
\[\begin{array}{lll}\int {h}^{'}(t)\ dt & = & \int 2.8369{e}^{-11.7096t}-0.8369dt \\ h(t) & = & -\frac{2.8369}{11.7096}{e}^{-11.7096t}-0.8369t+C \\ h(t) & = & -0.2423{e}^{-11.7096t}-0.8369t+C.\end{array}\]
Solve for \(C\) by using the initial condition:
\[\begin{array}{lll}h(t) & = & -0.2423{e}^{-11.7096t}-0.8369t+C \\ h(0) & = & -0.2423{e}^{-11.7096(0)}-0.8369(0)+C \\ 1 & = & -0.2423+C \\ C & = & 1.2423.\end{array}\]
Therefore
\[h(t)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423.\]
After \(0.104\) second, the height is given by
\(h(0.104)=-0.2423{e}^{-11.7096t}-0.8369t+1.2423\approx 1.0836\) meter.
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The weight of a penny is \(2.5\) grams (United States Mint, “Coin Specifications,” accessed April 9, 2015, http://www.usmint.gov/about_the_mint/?action=coin_specifications), and the upper observation deck of the Empire State Building is \(369\) meters above the street. Since the penny is a small and relatively smooth object, air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to \(0.0025v.\) Furthermore, the penny is dropped with no initial velocity imparted to it.
- Set up an initial-value problem that represents the falling penny.
- Solve the problem for \(v(t).\)
- What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as \(t\) approaches infinity)?
जवाब दिखाएँ
- \(\begin{array}{lll}\frac{dv}{dt} & = & \text{-}v-9.8 \\ v(0) & = & 0\end{array}\)
- \(v(t)=9.8({e}^{\text{-}t}-1)\)
- \(\underset{t\to \infty }{\text{lim}}v(t)=\underset{t\to \infty }{\text{lim}}(9.8({e}^{\text{-}t}-1))=-9.8\ \text{m/s}\approx -21.922\ \text{mph}\)
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A circuit has in series an electromotive force given by \(E=50\ \text{sin}\ 20t\ \text{V},\) a resistor of \(5\text{Ω},\) and an inductor of \(0.4\ \text{H}\text{.}\) If the initial current is \(0,\) find the current at time \(t>0.\)
जवाब दिखाएँ
We have a resistor and an inductor in the circuit, so we use . The voltage drop across the resistor is given by \({E}_{R}=Ri=5i.\) The voltage drop across the inductor is given by \({E}_{L}=L{i}^{'}=0.4{i}^{'}.\) The electromotive force becomes the right-hand side of . Therefore becomes
\[0.4{i}^{'}+5i=50\ \text{sin}\ 20t.\]Dividing both sides by \(0.4\) gives the equation
\[{i}^{'}+12.5i=125\ \text{sin}\ 20t.\]Since the initial current is 0, this result gives an initial condition of \(i(0)=0.\) We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.
Step 1. Rewrite the differential equation as \({i}^{'}+12.5i=125\ \text{sin}\ 20t.\) This gives \(p(t)=12.5\) and \(q(t)=125\ \text{sin}\ 20t.\)
Step 2. The integrating factor is \(\mu (t)={e}^{\int 12.5dt}={e}^{12.5t}.\)
Step 3. Multiply the differential equation by \(\mu (t)\text{:}\)
\[\begin{array}{lll}{e}^{12.5t}{i}^{'}+12.5{e}^{12.5t}i & = & 125{e}^{12.5t}\text{sin}\ 20t \\ \frac{d}{dt}\ [i{e}^{12.5t}] & = & 125{e}^{12.5t}\text{sin}\ 20t.\end{array}\]Step 4. Integrate both sides:
\[\begin{array}{lll}\int \frac{d}{dt}\ [i{e}^{12.5t}]\ dt & = & \int 125{e}^{12.5t}\text{sin}\ 20t\ dt \\ i{e}^{12.5t} & = & (\frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t}{89})\ {e}^{12.5t}+C \\ i(t) & = & \frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t}{89}+C{e}^{-12.5t}.\end{array}\]Step 5. Solve for \(C\) using the initial condition \(v(0)=2\text{:}\)
\[\begin{array}{lll}i(t) & = & \frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t}{89}+C{e}^{-12.5t} \\ i(0) & = & \frac{250\ \text{sin}\ 20(0)-400\ \text{cos}\ 20(0)}{89}+C{e}^{-12.5(0)} \\ 0 & = & -\frac{400}{89}+C \\ C & = & \frac{400}{89}.\end{array}\]Therefore the solution to the initial-value problem is \(i(t)=\frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t+400{e}^{-12.5t}}{89}=\frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t}{89}+\frac{400{e}^{-12.5t}}{89}.\)
The first term can be rewritten as a single cosine function. First, multiply and divide by \(\sqrt{{250}^{2}+{400}^{2}}=50\sqrt{89}\text{:}\)
\[\begin{array}{ll}\frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t}{89} & =\frac{50\sqrt{89}}{89}(\frac{250\ \text{sin}\ 20t-400\ \text{cos}\ 20t}{50\sqrt{89}}) \\ & =-\frac{50\sqrt{89}}{89}(\frac{8\ \text{cos}\ 20t}{\sqrt{89}}-\frac{5\ \text{sin}\ 20t}{\sqrt{89}}).\end{array}\]Next, define \(\phi\) to be an acute angle such that \(\text{cos}\ \phi =\frac{8}{\sqrt{89}}.\) Then \(\text{sin}\ \phi =\frac{5}{\sqrt{89}}\) and
\[\begin{array}{ll}-\frac{50\sqrt{89}}{89}(\frac{8\ \text{cos}\ 20t}{\sqrt{89}}-\frac{5\ \text{sin}\ 20t}{\sqrt{89}}) & =-\frac{50\sqrt{89}}{89}\ (\text{cos}\ \phi \ \text{cos}\ 20t-\text{sin}\ \phi \ \text{sin}\ 20t) \\ & =-\frac{50\sqrt{89}}{89}\text{cos}(20t+\phi ).\end{array}\]Therefore the solution can be written as
\[i(t)=-\frac{50\sqrt{89}}{89}\ \text{cos}\ (20t+\phi )+\frac{400{e}^{-12.5t}}{89}.\]The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is given by \(\phi ,\) and the amplitude of the steady-state current is given by \(\frac{50\sqrt{89}}{89}.\) The graph of this solution appears in :
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A circuit has in series an electromotive force given by \(E=20\ \text{sin}\ 5t\) V, a capacitor with capacitance \(0.02\ \text{F},\) and a resistor of \(8\ \text{Ω}.\) If the initial charge is \(4\ \text{C},\) find the charge at time \(t>0.\)
जवाब दिखाएँ
Initial-value problem:
\(8{q}^{'}+\frac{1}{0.02}q=20\ \text{sin}\ 5t,\ q(0)=4\)
\(q(t)=\frac{10\ \text{sin}\ 5t-8\ \text{cos}\ 5t+172{e}^{-6.25t}}{41}\)
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\(\frac{dy}{dx}={x}^{2}y+\text{sin}\ x\)
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\(\frac{dy}{dt}=ty\)
जवाब दिखाएँ
Yes
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\(\frac{dy}{dt}+{y}^{2}=x\)
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\(y'={x}^{3}+{e}^{x}\)
जवाब दिखाएँ
Yes
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\(y'=y+{e}^{y}\)
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\(y'={x}^{3}y+\text{sin}\ x\)
जवाब दिखाएँ
\(y'-{x}^{3}y=\text{sin}\ x\)
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\(y'+3y-\text{ln}\ x=0\)
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\(\text{-}xy'=(3x+2)y+x{e}^{x}\)
जवाब दिखाएँ
\(y'+\frac{(3x+2)}{x}y=\text{-}{e}^{x}\)
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\(\frac{dy}{dt}=4y+ty+\text{tan}\ t\)
-
\(\frac{dy}{dt}=yx(x+1)\)
जवाब दिखाएँ
\(\frac{dy}{dt}-yx(x+1)=0\)
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\(y'+{e}^{x}y=\text{sin}\ x\)
जवाब दिखाएँ
\({e}^{\left({e}^{x}\right)}\)
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\(y'=x\ \text{ln}(x)y+3x\)
-
\(\frac{dy}{dx}=\text{tanh}(x)y+1\)
जवाब दिखाएँ
\(\text{-}\text{ln}(\text{cosh}\ x)\)
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\(\frac{dy}{dt}+3ty={e}^{t}y\)
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\(y'=2y-{x}^{2}\)
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\(xy'=3y-6{x}^{2}\)
जवाब दिखाएँ
\(y=C{x}^{3}+6{x}^{2}\)
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\((x+2)y'=3x+y\)
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\(y'=3x+xy\)
जवाब दिखाएँ
\(y=C{e}^{{x}^{2}\text{/}2}-3\)
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\(\text{sin}(x)y'=y+2x\)
जवाब दिखाएँ
\(y=C\ \text{tan}(\frac{x}{2})-2x+4\ \text{tan}(\frac{x}{2})\text{ln}(\text{sin}(\frac{x}{2}))\)
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\(y'=y+{e}^{x}\)
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\(xy'=3y+{x}^{2}\)
जवाब दिखाएँ
\(y=C{x}^{3}-{x}^{2}\)
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\(y'+\text{ln}\ x=\frac{y}{x}\)
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[T] \((x+2)y'=2y-1\)
जवाब दिखाएँ
\(y=C{(x+2)}^{2}+\frac{1}{2}\)
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[T] \(y'=3{e}^{t\text{/}3}-2y\)
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[T] \(xy'+\frac{y}{2}=\text{sin}(3t)\)
जवाब दिखाएँ
\(y=\frac{C}{\sqrt{x}}+2\ \text{sin}(3t)\)
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[T] \(xy'=2\frac{\text{cos}\ x}{x}-3y\)
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[T] \((x+1)y'=3y+{x}^{2}+2x+1\)
जवाब दिखाएँ
\(y=C{(x+1)}^{3}-{x}^{2}-2x-1\)
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[T] \(\text{sin}(x)y'+\text{cos}(x)y=2x\)
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[T] \(\sqrt{{x}^{2}+1}y'=y+2\)
जवाब दिखाएँ
\(y=C{e}^{{\text{sinh}}^{-1}x}-2\)
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[T] \({x}^{3}y'+2{x}^{2}y=x+1\)
Symbols used here
Antiderivative (indefinite) or signed area from a to b (definite).
Average of the data; average of the whole population.
Instantaneous rate of change; slope of the graph.
Prime notation for derivatives with respect to x (or t).
i² = −1.
The two sides are different.
Ratio of a circle's circumference to its diameter, 3.14159…
2.71828…, the base whose exponential is its own derivative.
Not a number: "grows without bound" in limits and intervals.
Ratios of sides in a right triangle; coordinates on the unit circle.
The exponent b must be raised to for x; ln uses base e.
Add a_k for k = 1 up to n.
The value f(x) approaches as x approaches a.
Constants of integration fixed by initial conditions.
How to: First-Order Linear Equations
- Write a first-order linear differential equation in standard form.
- Find an integrating factor and use it to solve a first-order linear differential equation.
- Solve applied problems involving first-order linear differential equations.
- Add
- Multiply both sides by
- Subtract
- Put the equation into standard form and identify
- Calculate the integrating factor
Questions people ask
What is a derivative in one sentence?
The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.
अपने आप की कोशिश करें
Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
अधिक में Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests