Definitions [13]
If an equation contains derivatives of one dependent variable with respect to one or more independent variables, then it is called a differential equation.
Example
This is a differential equation because it contains the derivative \[\frac{dy}{dx}\].
A differential equation that contains ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation.
Example:
The highest exponent of the highest derivative is called the degree of a differential equation, provided exponents of each derivative and an unknown variable appearing in the differential equation are non-negative integers.
The order of the highest differential coefficient (or the highest order derivative appearing in a differential equation) is the order of the differential equation.
A solution obtained from the general solution by assigning specific values to the arbitrary constants is called a particular solution.
For a differential equation, a solution is a function that makes the left-hand side equal to the right-hand side when the function and its required derivatives are substituted. If y = ϕ(x) satisfies the differential equation, then the curve represented by y = ϕ(x) is called the solution curve or integral curve.
A solution containing arbitrary constants is called the general solution of a differential equation.
The equation \[\frac{dy}{dx} = F(x, y)\] is said to be in variable separable form if it can be expressed as \[g(x) dx = h(y) dy\] or equivalently as \[\frac{dy}{dx} = g(x)h(y)\] so that the variables can be separated and integrated.
A differential equation of the form \[\frac{dy}{dx}=\frac{f_{1}(x,y)}{f_{2}(x,y)},\] where f1(x, y) ) and f2(x, y) are homogeneous functions of x and y of the same degree, is called a homogeneous differential equation.
A linear differential equation of first order and first degree is
\[\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{P}y=\mathrm{Q}\], where P and Q are the functions of x or constants. Its general solution is \[y.\left(\mathrm{I.F.}\right)=\int\mathrm{Q.}\left(\mathrm{I.F.}\right)\mathrm{d}x+\mathrm{c}\] and the function \[\mathrm{e}^{\int\mathrm{Pdx}}\] is called the integrating factor (I.F.) of the given equation.
The factor e∫P dx on multiplying by which the left-hand side of the differential equation\[\frac{dy}{dx}+Py=Q\] becomes the differential coefficient of a function of x and y, is called the integrating factor of the differential equation.
General Solution: \[y\cdot\mathrm{I.F.}=\int Q\cdot\mathrm{I.F.}dx+c\]
A differential equation is non-linear if any one of the following holds:
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The degree is more than one
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Any differential coefficient has an exponent of more than one
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Exponent of the dependent variable is more than one
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Products containing the dependent variable and its differential coefficients are present
A first-order differential equation, along with an initial condition, is called an initial value problem.
Theorems and Laws [1]
Prove that:
`int_0^(2a)f(x)dx = int_0^af(x)dx + int_0^af(2a - x)dx`
Since ‘a’ lies between 0 and 2a,
we have
`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx, .......(byint_a^bf(x)dx=int_a^cf(x)dx+int_c^bf(x)dx)`
`=I_1+I_2` ........................(say)
`I_2 = int_a^(2a)f(x)dx`
Put x = 2a − t
Therefore, dx = −dt
When x = a, 2a − t = a
t = a
When x = 2a, 2a − t = 2a
t = 0
`I_2 = int_0^(2a) f(x) dx = int_a^0 f(2a - t) (-dt)`
`= -int_a^0 f(2a - t)dt = int_0^a f(2a - t)dt ...................... (By int_a^b f(x)dx = -int_b^a f(x)dx)`
`=int_0^a f(2a - x)dx ..............(By int_a^b f(X)dx = int_a^b f(t)dt)`
`int_0^(2a) f(x)dx = int_0^a f(x)dx + int_0^a f(2a - x)dx`
`= int_0^a [f(x) + f(2a - x)]dx`
To show that:
`int_0^pi sin x dx = 2 int_0^(pi/2) sin x dx`
We use the proven property by setting f(x) = sin x and 2a = π, which means a = `pi/2`.
The property tells us that:
`int_0^pi sin x dx = int_0^(pi/2) sin x dx + int_0^(pi/2) sin (pi - x) dx`
Knowing the trigonometric identity sin (π - x) = sin x, the equation simplifies to:
`int_0^pi sin x dx = 2 int_0^(pi/2) sin x dx`
This directly applies the property to the integral of sin x over [0, π] to show it equals twice the integral of sin x over `[0, pi/2]`, demonstrating the utility of this property in simplifying integrals with symmetric functions over specific intervals.
Key Points
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A differential equation contains derivatives.
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An ordinary differential equation contains derivatives with respect to only one independent variable.
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Differential equations describe rates of change in mathematics and science.
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Order = highest derivative order.
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Degree = power of highest derivative.
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Degree exists only for polynomial equations in derivatives.
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Always check polynomial condition before stating the degree.
- A differential equation contains derivatives of an unknown function.
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Its solution is generally a function, not a single number.
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The graph of the solution function is called the solution curve or integral curve.
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A general solution contains arbitrary constants.
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A particular solution is obtained by assigning fixed values to those constants.
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To verify a solution, substitute the function and its derivatives into the equation and check whether LHS = RHS.
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Variable separable equations can be rewritten as x-part = y-part.
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Separate variables first, then integrate.
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Use one constant of integration.
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Apply the initial condition only after getting the general solution.
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Final answers may be explicit or implicit.
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Check homogeneity first.
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Differentiate substitution carefully.
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Convert to separable form.
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Back-substitute to original variables.
- Write the equation in the form dy/dx + Py = Q
- Identify P and Q
- Find I.F. = \[\mathrm{e}^{\int\mathrm{Pdx}}\]
- Multiply the whole equation by I.F.
- Integrate and get a solution.
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Write the equation in the form
\[\frac{dy}{dx}+Py=Q\] -
Find the integrating factor
\[\mathrm{I.F.}=e^{\int Pdx}\] -
Multiply the entire equation by I.F.
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Integrate both sides w.r.t x
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Obtain
\[y(\mathrm{I.F.})=\int Q(\mathrm{I.F.})dx+c\]
- Radioactive Decay: \[x=x_0e^{-kt}\]
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Half-Life Formula: \[k=\frac{\ln2}{T}\]
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Newton’s Law of Cooling: \[\theta=\theta_0+(\theta_1-\theta_0)e^{-kt}\]
- Population Growth: \[P=ae^{kt}\]
Important Questions [57]
- Solve the Following Differential Equation: Cosec X Log Y D Y D X + X 2 Y 2 = 0
- Solve the Following Differential Equation : ( √ 1 + X 2 + Y 2 + X 2 Y 2 ) D X + X Y D Y = 0 .
- Solve the Following Differential Equation : Y 2 D X + ( X 2 − X Y + Y 2 ) D Y = 0 .
- If xmyn = (x + y)m+n, prove that dydx=yx.
- Find the particular solution of the differential equation "dy"/"dx" = "xy"/("x"^2+"y"^2),given that y = 1 when x = 0
- The Integrating Factor of the Differential Equation ( 1 − Y 2 ) D X D Y + Y X = a Y ( − 1 < Y < 1 ) is
- Solve the equation for x: sin^(−1) 5/x + sin^(−1) 12/x = π/2, x ≠ 0
- Find the general solution of the following differential equation: dydx=ex-y+x2e-y
- The sum of the order and the degree of the differential equation ddx[(dydx)3] is ______.
- Write the Order and Degree of the Differential Equation ( D 4 Y D X 4 ) 2 = X + ( D Y D X ) 2 − 3
- The order and the degree of the differential equation (1+3dydx)2=4d3ydx3 respectively are ______.
- Degree of the differential equation sinx+cos(dydx) = y2 is ______.
- Write the Sum of the Order and Degree of the Differential Equation ( D 2 Y D X 2 ) 2 + ( D Y D X ) 3 + X 4 = 0 .
- Find the Order and the Degree of the Differential Equation X 2 D 2 Y D X 2 = { 1 + ( D Y D X ) 2 } 4
- Write the Order and the Degree of the Following Differential Equation: X 3 ( D 2 Y D X 2 ) 2 + X ( D Y D X ) 4 = 0
- Write the degree of the differential equation x^3((d^2y)/(dx^2))^2+x(dy/dx)^4=0
- Find the Particular Solution of the Differential Equation ( 1 + Y 2 ) + ( X − E Tan − 1 Y ) D Y D X = Given that Y = 0 When X = 1.
- Write the Solution of the Differential Equation D Y D X = 2 − Y .
- Find the Particular Solution of the Differential Equation D Y D X = X ( 2 Log X + 1 ) Sin Y + Y Cos Y Given that Y = π 2 When X = 1.
- Solve the Differential Equation (X2 − Yx2) Dy + (Y2 + X2y2) Dx = 0, Given that Y = 1, When X = 1.
- Find the particular solution of the differential equation e^x √(1−y^2)dx+y/x dy=0 , given that y=1 when x=0
- Solve the differential equation: (1 + y2) dx = (tan−1 y − x) dy
- Form the Differential Equation of the Family of Circles in the Second Quadrant and Touching the Coordinate Axes.
- Find the differential equation representing the curve y = cx + c2.
- Find the particular solution of differential equation: dy/dx=(−x+ycosx)/(1+sinx) given that y=1 when x=0
- Find the particular solution of the differential equation (1 – y^2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.
- Find the general solution of the following differential equation : (1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0
- Find the particular solution of the differential equation (1+x^2)dy/dx=(e^(m tan^-1 x)-y), give that y=1 when x=0.
- Find the particular solution of the differential equation dy/dx=(xy)/(x^2+y^2) given that y = 1, when x = 0.
- If y = P eax + Q ebx, show that dy/dx^2=(a+b)dy/dx + aby=0
- Find the particular solution of the differential equation dy/dx=1 + x + y + xy, given that y = 0 when x = 1.
- Find the particular solution of the differential equation log(dy/dx)= 3x + 4y, given that y = 0 when x = 0.
- Find the particular solution of the differential equation x (1 + y2) dx – y (1 + x2) dy = 0, given that y = 1 when x = 0.
- Find (Dy)By(Dx) at X = 1, Y = `Piby4` If `Sin^2 Y + Cos Xy = K`
- If `Y = Sin Power (-1) (6xsquaeroot(1-9x^2))`, `1by(3squareroot2) < X < 1/(3squarroott2)` Then Find `(Dy)By(Dx)`
- Find the Particular Solution of the Differential Equation `Tan X (Dy)By(Dx) = 2x Tan X + X^2 - Y`; `(Tan X Not Equal 0)` Given that Y = 0 When `X
- Find the General Solution of the Differential Equation X Cos ( Y X ) D Y D X = Y Cos ( Y X ) + X .
- Solve the Differential Equation: D Y D X − 2 X 1 + X 2 Y' = X 2 + 2
- Solve the Differential Equation: ( X + 1 ) D Y D X = 2 E − Y − 1 ; ( 0 ) = 0 .
- Solve the Differential Equation : ( X 2 + 3 Xy + Y 2 ) D X − X 2 D Y = 0 Given that Y = 0 When X = 1 .
- Find the general solution of y2dx + (x2 – xy + y2) dy = 0.
- Solve: 2(y + 3) - xy (dy)/(dx) = 0, given that y(1) = – 2.
- Find the particular solution of the differential equation xdydx-y=x2.ex, given y(1) = 0.
- Find the general solution of the differential equation xdydx=y(logy-logx+1).
- Find the general solution of the differential equation: log(dydx)=ax+by.
- Find the general solution of the differential equation: dydx=3e2x+3e4xex+e-x
- Show that the differential equation 2xy dy/dx=x^2+3y^2 is homogeneous and solve it.
- Find the particular solution of the differential equation: 2y ex/y dx + (y - 2x ex/y) dy = 0 given that x = 0 when y = 1.
- Solve the Differential Equation: D Y D X = X + Y X − Y
- Solve the Following Differential Equation : [ Y − X Cos ( Y X ) ] D Y + [ Y Cos ( Y X ) − 2 X Sin ( Y X ) ] D X = 0
- Find the general solution of the differential equation: (xy – x2) dy = y2 dx
- Solve the Differential Equation
- Read the following passage: An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation.
- Prove that X2 – Y2 = C(X2 + Y2)2 is the General Solution of the Differential Equation (X3 – 3xy2)Dx = (Y3 – 3x2y)Dy, Where C is Parameter
- Find the Particular Solution of the Differential Equation `(X - Y) Dy/Dx = (X + 2y)` Given that Y = 0 When X = 1.
- Solve the Differential Equation: X Dy - Y Dx = √ X 2 + Y 2 D X , Given that Y = 0 When X = 1.
- Show that the differential equation 2y^(x/y) dx + (y − 2x e^(x/y)) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.
Concepts [7]
- Basic Concepts of Differential Equations
- Order and Degree of a Differential Equation
- General and Particular Solutions of a Differential Equation
- Methods of Solving Differential Equations> Variable Separable Differential Equations
- Methods of Solving Differential Equations> Homogeneous Differential Equations
- Methods of Solving Differential Equations>Linear Differential Equations
- Overview of Differential Equations
