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Differentiating a function, A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. Example Consider the straight line y = 3x + 2 We can calculate the gradient of this line as follows. We take two points and calculate the change in y divided by the change in x. When x changes from −1 to 0, y changes from −1 to 2, and so No matter which pair of points we choose the value of the gradient is always 3. y is a function of x, so y =f(x) f(x1) = y1 and f(x2) =y2 x2-x1=dx, x2= x1+dx, By substitituting these terms in slope formula, we have; m =[ f(x1 + dx) - f(x1)]/dx, as the limit of dx approaches 0 Seven steps to avoit the spread of Corona Virus (Covid19): If you find this video interesting, kindly subscribe to my channel for more exciting Maths tutorials. Subscribe link: #Differentiation #First #Principles WhatsApp group: Facebook: Instagram: Linkedin: Blog:

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IMVU_jxt_•Oct 22, 2025

Integration by partial fractions is an integration technique which uses partial fraction decomposition to simplify the integrand. The integrand is written as partial fractions and then evaluated using standard methods. The integrals of many rational functions lead to a natural log function with absolute value expressions. This video explains what to do when you have non repeated linear factors factors.

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Integration by partial fractions is an integration technique which uses partial fraction decomposition to simplify the integrand. The integrand is written as partial fractions and then evaluated using standard methods. The integrals of many rational functions lead to a natural log function with absolute value expressions. This video explains what to do when you have non repeated linear factors factors.

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To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

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To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

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Consider a differential equation of type y′′+py′+qy=0, where p,q are some constant coefficients. For each of the equation we can write the so-called characteristic (auxiliary) equation: k2+pk+q=0. The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options: Discriminant of the characteristic quadratic equation D is greater than 0. Then the roots of the characteristic equations r1 and r2 are real and distinct. In this case the general solution is given by the following function y(x)=C₁eʳ¹ˣ+C₂eʳ²ˣ, where C1 and C2 are arbitrary real numbers. Discriminant of the characteristic quadratic equation D=0. Then the roots are real and equal. It is said in this case that there exists one repeated root r of order 2. The general solution of the differential equation has the form: y(x)=(C₁x+C₂)eʳˣ. Discriminant of the characteristic quadratic equation D is less than 0. Such an equation has complex roots

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Consider a differential equation of type y′′+py′+qy=0, where p,q are some constant coefficients. For each of the equation we can write the so-called characteristic (auxiliary) equation: k2+pk+q=0. The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options: Discriminant of the characteristic quadratic equation D is greater than 0. Then the roots of the characteristic equations r1 and r2 are real and distinct. In this case the general solution is given by the following function y(x)=C₁eʳ¹ˣ+C₂eʳ²ˣ, where C1 and C2 are arbitrary real numbers. Discriminant of the characteristic quadratic equation D=0. Then the roots are real and equal. It is said in this case that there exists one repeated root r of order 2. The general solution of the differential equation has the form: y(x)=(C₁x+C₂)eʳˣ. Discriminant of the characteristic quadratic equation D is less than 0. Such an equation has complex roots

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To solve any Second Order Linear Homogeneous Differential Equation, first this you need to do, is to transform the equation in to an auxiliary or characteristics equation in the form: ar²+be+c=0 We have already seen how to do that in our previous lesson. The next move is to solve for r which are the roots of the equation (r intercept). Then determine the nature of the roots and substitute in to the following equations, depending on the nature of roots. y=C₁eʳ¹ˣ+C₂eʳ²ˣ. when you obtain real and distinct roots y=(C₁+C₂x)eʳˣ when you Obtain real and equal roots. and finally, if you Obtain a complex solution in the form: r= m+si or r = m-si where i is imaginary number, and m and s are real numbers, then y=eᵐˣ[C₁cos(st)+C₂sin(st)]

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Happy Ney Year 2021 Mathematics #2021 #HappyNewYear #Mathematics

James ReidOct 22, 2025

Happy Ney Year 2021 Mathematics #2021 #HappyNewYear #Mathematics