Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form:<\/p>\r\n
[latex]ay^{\\prime\\prime}+by^\\prime+cy=0[\/latex]<\/p>\r\nwhere [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are constants.\r\n
Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let\u2019s see what happens when we try a solution of the form [latex]y(x)=e^{\\lambda x}[\/latex], where [latex]\\lambda[\/latex] (the lowercase Greek letter lambda) is some constant.<\/p>\r\n
If [latex]y(x)=e^{\\lambda x}[\/latex], then [latex]y^\\prime(x)=\\lambda e^{\\lambda x}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\lambda^2e^{\\lambda x}[\/latex]. Substituting these expressions into a linear second-order differential equation, we get<\/p>\r\n
[latex]\\begin{aligned}\r\nay^{\\prime\\prime}+by^\\prime+cy&=a(\\lambda^2e^{\\lambda x})+b(\\lambda e^{\\lambda x})+ce^{\\lambda x} \\\\\r\n&=e^{\\lambda x}(a\\lambda^2+b\\lambda+c)\r\n\\end{aligned}[\/latex].<\/p>\r\n
Since [latex]e^{\\lambda x}[\/latex] is never zero, this expression can be equal to zero for all [latex]x[\/latex] only if:<\/p>\r\n
[latex]\\large{a\\lambda^1+b\\lambda+c=0}[\/latex].<\/p>\r\n
We call this the characteristic equation<\/strong> of the differential equation.<\/p>\r\n\r\n The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula:<\/p>\r\n [latex]\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}[\/latex].<\/p>\r\n This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.<\/p>\r\n\r\n [latex]\\large{y(x)=c_1e^{\\lambda_1 x}+c_2e^{\\lambda_2 x}}[\/latex],<\/span><\/p>\r\n where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n\r\n The differential equation [latex]y^{\\prime\\prime}9y^\\prime+14y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+9\\lambda+14=0[\/latex]. This factors into [latex](\\lambda+2)(\\lambda+7)=0[\/latex], which has roots [latex]\\lambda_1=-2[\/latex] and [latex]\\lambda_2=-7[\/latex]. Therefore, the general solution to this differential equation is:<\/p>\r\n [latex]y(x)=c_1e^{-2}+c_2e^{-7x}[\/latex].<\/p>\r\n\r\n<\/section><\/section> [latex]\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}[\/latex].<\/p>\r\n But, [latex]\\lambda[\/latex] is a repeated root, so [latex]b^2-4ac=0[\/latex] and [latex]\\lambda=\\frac{-b}{2a}[\/latex]. Thus, if [latex]y=xe^{\\lambda x}[\/latex], we have:<\/p>\r\n [latex]y^\\prime=e^{\\lambda x}+\\lambda xe^{\\lambda x}\\text{ and }y^{\\prime\\prime}=2\\lambda e^{\\lambda x}+\\lambda^2xe^{\\lambda x}[\/latex].<\/p>\r\n Substituting these expressions into a linear second-order differential equation, we see that:<\/p>\r\n [latex]\\begin{aligned}\r\nay^{\\prime\\prime}+by^\\prime+cy&=a(2\\lambda e^{\\lambda x}+\\lambda^2e^{\\lambda x})+b(e^{\\lambda x}+\\lambda xe^{\\lambda x})+cxe^{\\lambda x} \\\\\r\n&=xe^{\\lambda x}(a\\lambda^2+b\\lambda+c)+e^{\\lambda x}(2a\\lambda+b) \\\\\r\n&=xe^{\\lambda x}(0)+e^{\\lambda x}\\left(2\\left(\\frac{-b}{2a}\\right)+b\\right) \\\\\r\n&=0+e^{\\lambda x}(0) \\\\\r\n=0\r\n\\end{aligned}[\/latex].<\/p>\r\nThis shows that [latex]xe^{\\lambda x}[\/latex] is a solution to a linear second-order differential equation. Since [latex]e^{\\lambda x}[\/latex] and [latex]xe^{\\lambda x}[\/latex] are linearly independent, when the characteristic equation has a repeated root [latex]\\lambda[\/latex], the general solution to a linear second-order differential equation is given by\r\n [latex]\\large{y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}}[\/latex],<\/p>\r\n where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n\r\n The differential equation [latex]y^{\\prime\\prime}+12y^\\prime+36y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+12\\lambda+36=0[\/latex]. This factors into [latex](\\lambda+6)^2[\/latex], which has a repeated root [latex]\\lambda=-6[\/latex]. Therefore, the general solution to this differential equation is:<\/p>\r\n\r\n The third case we must consider is when [latex]b^2-4ac<0[\/latex]. In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number [latex]i=\\sqrt{-1}[\/latex] to find the roots, which take the form [latex]\\lambda_1=\\alpha+\\beta i[\/latex] and [latex]\\lambda_2=\\alpha-\\beta i[\/latex]. The complex number [latex]\\alpha+\\beta i[\/latex] is called the conjugate of [latex]\\alpha-\\beta i[\/latex]. Thus, when [latex]b^2-4ac<0[\/latex], the roots of our characteristic equation are always complex conjugates.<\/p>\r\n This creates a problem for us. If we follow the same process we used for distinct real roots\u2014using the roots of the characteristic equation as the coefficients in the exponents of exponential functions\u2014we get the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-valued functions, we do not address the idea of a derivative for such functions here.<\/p>\r\nIf possible, we'd like to find two linearly independent real-valued solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as if they were real-valued functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-valued functions do not follow the same rules as real-valued functions.\r\n\r\n [latex]e^{i\\theta}=\\cos\\theta+i\\sin\\theta[\/latex]<\/p>\r\n for all real numbers [latex]\\theta[\/latex].<\/p>\r\n\r\n<\/section>\r\n Based on the roots [latex]\\alpha\\pm\\beta i[\/latex] of the characteristic equation, the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] are linearly independent solutions to the differential equation. and the general solution is given by:<\/p>\r\n [latex]y(x)=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x}[\/latex].<\/p>\r\n Using some smart choices for [latex]c_1[\/latex] and [latex]c_2[\/latex], and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to\u00a0a linear second-order differential equation\u00a0and express our general solution in those terms.<\/p>\r\n Going back to the general solution, we have:<\/p>\r\n [latex]\\begin{aligned}\r\ny(x)&=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x} \\\\\r\n&=c_1e^{\\alpha x}e^{\\beta ix}+c_2e^{\\alpha x}e^{-\\beta ix} \\\\\r\n&=e^{\\alpha x}\\left(c_1e^{\\beta ix}+c_2e^{-\\beta ix}\\right)\r\n\\end{aligned}[\/latex].<\/p>\r\n Applying Euler\u2019s formula together with the identities [latex]\\cos(-x)=\\cos x[\/latex] and [latex]\\sin(-x)=-\\sin x[\/latex], we get:<\/p>\r\n [latex]\\begin{aligned}\r\ny(x)&=e^{\\alpha x}[c_1(\\cos\\beta x+i\\sin\\beta x)+c_2(\\cos(-\\beta x)+i\\sin(-\\beta x))] \\\\\r\n&=e^{\\alpha x}[(c_1+c_2)\\cos\\beta x+(c_1-c_2)i\\sin\\beta x]\r\n\\end{aligned}[\/latex].<\/p>\r\n Now, if we choose [latex]c_1=c_2=\\frac{1}{2}[\/latex], the second term is zero and we get:<\/p>\r\n [latex]y(x)=e^{\\alpha x}\\cos\\beta x[\/latex]<\/p>\r\n as a real-value solution to a linear second-order differential equation. Similarly, if we choose [latex]c_1=-\\frac{i}2[\/latex] and [latex]c_2=\\frac{i}2[\/latex], the first term is zero and we get:<\/p>\r\n [latex]y(x)=e^{\\alpha x}\\sin\\beta x[\/latex]<\/p>\r\n as a second, linearly independent, real-value solution to\u00a0a linear second-order differential equation.<\/p>\r\n Based on this, we see that if the characteristic equation has complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex], then the general solution to a linear second-order differential equation is given by:<\/p>\r\n [latex]\\begin{aligned}\r\ny(x)&=c_1e^{\\alpha x}\\cos\\beta x+c_2e^{\\alpha x}\\cos\\sin\\beta x \\\\\r\n&=e^{\\alpha x}(c_1\\cos\\beta x+c_2\\sin\\beta x)\r\n\\end{aligned}[\/latex].<\/p>\r\n where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n\r\n The differential equation [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2-2\\lambda+5=0[\/latex]. By the quadratic formula, the roots of the characteristic equation are [latex]1\\pm2i[\/latex]. Therefore, the general solution to this differential equation is:<\/p>\r\n [latex]y(x)=e^x(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\n\r\n<\/section>\r\n Find the general solution to the following differential equations. Give your answers as functions of [latex]x[\/latex].<\/p>\r\n\r\n Note that all these equations are already given in standard form (step 1).<\/p>\r\n\r\n [latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex] (step 4).<\/p>\r\n<\/li>\r\n \t [latex]y(t)=e^{-3t}(c_1\\cos2t+c_2\\sin2t)[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t [latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t [latex]y(x)=c_1+c_2e^{5x}[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t [latex]y(x)=c_1e^{4x}+c_2e^{-4x}[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t [latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex] (step 4).<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":" Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form:<\/p>\n [latex]ay^{\\prime\\prime}+by^\\prime+cy=0[\/latex]<\/p>\n where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are constants.<\/p>\n Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let\u2019s see what happens when we try a solution of the form [latex]y(x)=e^{\\lambda x}[\/latex], where [latex]\\lambda[\/latex] (the lowercase Greek letter lambda) is some constant.<\/p>\n If [latex]y(x)=e^{\\lambda x}[\/latex], then [latex]y^\\prime(x)=\\lambda e^{\\lambda x}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\lambda^2e^{\\lambda x}[\/latex]. Substituting these expressions into a linear second-order differential equation, we get<\/p>\n [latex]\\begin{aligned} ay^{\\prime\\prime}+by^\\prime+cy&=a(\\lambda^2e^{\\lambda x})+b(\\lambda e^{\\lambda x})+ce^{\\lambda x} \\\\ &=e^{\\lambda x}(a\\lambda^2+b\\lambda+c) \\end{aligned}[\/latex].<\/p>\n Since [latex]e^{\\lambda x}[\/latex] is never zero, this expression can be equal to zero for all [latex]x[\/latex] only if:<\/p>\n [latex]\\large{a\\lambda^1+b\\lambda+c=0}[\/latex].<\/p>\n We call this the characteristic equation<\/strong> of the differential equation.<\/p>\n The\u00a0characteristic equation<\/span><\/strong>\u00a0of the differential equation [latex]ay^{\\prime\\prime}+by^\\prime+cy=0[\/latex] is [latex]a\\lambda^2+b\\lambda+c+0[\/latex].<\/p>\n<\/section>\n The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula:<\/p>\n [latex]\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}[\/latex].<\/p>\n This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.<\/p>\n If the characteristic equation has distinct real roots [latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex], then [latex]e^{\\lambda_1 x}[\/latex] and [latex]e^{\\lambda_2 x}[\/latex] are linearly independent solutions to the differential equation, and the general solution is given by:<\/p>\n [latex]\\large{y(x)=c_1e^{\\lambda_1 x}+c_2e^{\\lambda_2 x}}[\/latex],<\/span><\/p>\n where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n The differential equation [latex]y^{\\prime\\prime}9y^\\prime+14y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+9\\lambda+14=0[\/latex]. This factors into [latex](\\lambda+2)(\\lambda+7)=0[\/latex], which has roots [latex]\\lambda_1=-2[\/latex] and [latex]\\lambda_2=-7[\/latex]. Therefore, the general solution to this differential equation is:<\/p>\n [latex]y(x)=c_1e^{-2}+c_2e^{-7x}[\/latex].<\/p>\n<\/section>\n<\/section>\n Things are a little more complicated if the characteristic equation has a repeated real root, [latex]\\lambda[\/latex]. In this case, we know [latex]e^{\\lambda x}[\/latex] is a solution to the differential equation, but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form [latex]ke^{\\lambda x}[\/latex], where [latex]k[\/latex] is some constant, but it would not be linearly independent of [latex]e^{\\lambda x}[\/latex]. Therefore, let’s try [latex]xe^{\\lambda x}[\/latex] as the second solution.<\/p>\n First, note that by the quadratic formula,<\/p>\n [latex]\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}[\/latex].<\/p>\n But, [latex]\\lambda[\/latex] is a repeated root, so [latex]b^2-4ac=0[\/latex] and [latex]\\lambda=\\frac{-b}{2a}[\/latex]. Thus, if [latex]y=xe^{\\lambda x}[\/latex], we have:<\/p>\n [latex]y^\\prime=e^{\\lambda x}+\\lambda xe^{\\lambda x}\\text{ and }y^{\\prime\\prime}=2\\lambda e^{\\lambda x}+\\lambda^2xe^{\\lambda x}[\/latex].<\/p>\n Substituting these expressions into a linear second-order differential equation, we see that:<\/p>\n [latex]\\begin{aligned} ay^{\\prime\\prime}+by^\\prime+cy&=a(2\\lambda e^{\\lambda x}+\\lambda^2e^{\\lambda x})+b(e^{\\lambda x}+\\lambda xe^{\\lambda x})+cxe^{\\lambda x} \\\\ &=xe^{\\lambda x}(a\\lambda^2+b\\lambda+c)+e^{\\lambda x}(2a\\lambda+b) \\\\ &=xe^{\\lambda x}(0)+e^{\\lambda x}\\left(2\\left(\\frac{-b}{2a}\\right)+b\\right) \\\\ &=0+e^{\\lambda x}(0) \\\\ =0 \\end{aligned}[\/latex].<\/p>\n This shows that [latex]xe^{\\lambda x}[\/latex] is a solution to a linear second-order differential equation. Since [latex]e^{\\lambda x}[\/latex] and [latex]xe^{\\lambda x}[\/latex] are linearly independent, when the characteristic equation has a repeated root [latex]\\lambda[\/latex], the general solution to a linear second-order differential equation is given by<\/p>\n [latex]\\large{y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}}[\/latex],<\/p>\n where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n The differential equation [latex]y^{\\prime\\prime}+12y^\\prime+36y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+12\\lambda+36=0[\/latex]. This factors into [latex](\\lambda+6)^2[\/latex], which has a repeated root [latex]\\lambda=-6[\/latex]. Therefore, the general solution to this differential equation is:<\/p>\n The third case we must consider is when [latex]b^2-4ac<0[\/latex]. In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number [latex]i=\\sqrt{-1}[\/latex] to find the roots, which take the form [latex]\\lambda_1=\\alpha+\\beta i[\/latex] and [latex]\\lambda_2=\\alpha-\\beta i[\/latex]. The complex number [latex]\\alpha+\\beta i[\/latex] is called the conjugate of [latex]\\alpha-\\beta i[\/latex]. Thus, when [latex]b^2-4ac<0[\/latex], the roots of our characteristic equation are always complex conjugates.<\/p>\n This creates a problem for us. If we follow the same process we used for distinct real roots\u2014using the roots of the characteristic equation as the coefficients in the exponents of exponential functions\u2014we get the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-valued functions, we do not address the idea of a derivative for such functions here.<\/p>\n If possible, we’d like to find two linearly independent real-valued solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as if they were real-valued functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-valued functions do not follow the same rules as real-valued functions.<\/p>\n [latex]e^{i\\theta}=\\cos\\theta+i\\sin\\theta[\/latex]<\/p>\n for all real numbers [latex]\\theta[\/latex].<\/p>\n<\/section>\n Based on the roots [latex]\\alpha\\pm\\beta i[\/latex] of the characteristic equation, the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] are linearly independent solutions to the differential equation. and the general solution is given by:<\/p>\n [latex]y(x)=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x}[\/latex].<\/p>\n Using some smart choices for [latex]c_1[\/latex] and [latex]c_2[\/latex], and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to\u00a0a linear second-order differential equation\u00a0and express our general solution in those terms.<\/p>\n Going back to the general solution, we have:<\/p>\n [latex]\\begin{aligned} y(x)&=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x} \\\\ &=c_1e^{\\alpha x}e^{\\beta ix}+c_2e^{\\alpha x}e^{-\\beta ix} \\\\ &=e^{\\alpha x}\\left(c_1e^{\\beta ix}+c_2e^{-\\beta ix}\\right) \\end{aligned}[\/latex].<\/p>\n Applying Euler\u2019s formula together with the identities [latex]\\cos(-x)=\\cos x[\/latex] and [latex]\\sin(-x)=-\\sin x[\/latex], we get:<\/p>\n [latex]\\begin{aligned} y(x)&=e^{\\alpha x}[c_1(\\cos\\beta x+i\\sin\\beta x)+c_2(\\cos(-\\beta x)+i\\sin(-\\beta x))] \\\\ &=e^{\\alpha x}[(c_1+c_2)\\cos\\beta x+(c_1-c_2)i\\sin\\beta x] \\end{aligned}[\/latex].<\/p>\n Now, if we choose [latex]c_1=c_2=\\frac{1}{2}[\/latex], the second term is zero and we get:<\/p>\n [latex]y(x)=e^{\\alpha x}\\cos\\beta x[\/latex]<\/p>\n as a real-value solution to a linear second-order differential equation. Similarly, if we choose [latex]c_1=-\\frac{i}2[\/latex] and [latex]c_2=\\frac{i}2[\/latex], the first term is zero and we get:<\/p>\n [latex]y(x)=e^{\\alpha x}\\sin\\beta x[\/latex]<\/p>\n as a second, linearly independent, real-value solution to\u00a0a linear second-order differential equation.<\/p>\n Based on this, we see that if the characteristic equation has complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex], then the general solution to a linear second-order differential equation is given by:<\/p>\n [latex]\\begin{aligned} y(x)&=c_1e^{\\alpha x}\\cos\\beta x+c_2e^{\\alpha x}\\cos\\sin\\beta x \\\\ &=e^{\\alpha x}(c_1\\cos\\beta x+c_2\\sin\\beta x) \\end{aligned}[\/latex].<\/p>\n where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n The differential equation [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2-2\\lambda+5=0[\/latex]. By the quadratic formula, the roots of the characteristic equation are [latex]1\\pm2i[\/latex]. Therefore, the general solution to this differential equation is:<\/p>\n [latex]y(x)=e^x(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\n<\/section>\n We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.<\/p>\n Find the general solution to the following differential equations. Give your answers as functions of [latex]x[\/latex].<\/p>\ncharacteristic equation<\/h3>\r\nThe\u00a0characteristic equation<\/span><\/strong>\u00a0of the differential equation [latex]ay^{\\prime\\prime}+by^\\prime+cy=0[\/latex] is [latex]a\\lambda^2+b\\lambda+c+0[\/latex].\r\n\r\n<\/section>\r\n
Distinct Real Roots<\/h3>\r\nIf the characteristic equation has distinct real roots [latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex], then [latex]e^{\\lambda_1 x}[\/latex] and [latex]e^{\\lambda_2 x}[\/latex] are linearly independent solutions to the differential equation, and the general solution is given by:\r\n
Single Repeated Real Root<\/h3>\r\nThings are a little more complicated if the characteristic equation has a repeated real root, [latex]\\lambda[\/latex]. In this case, we know [latex]e^{\\lambda x}[\/latex] is a solution to the differential equation, but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form [latex]ke^{\\lambda x}[\/latex], where [latex]k[\/latex] is some constant, but it would not be linearly independent of [latex]e^{\\lambda x}[\/latex]. Therefore, let's try [latex]xe^{\\lambda x}[\/latex] as the second solution.\r\n\r\nFirst, note that by the quadratic formula,\r\n
Complex Conjugate Roots<\/h3>\r\n
Summary of Results<\/h3>\r\n
summary of characteristic equation cases<\/h3>\r\n
\r\n\r\n
\r\n \r\nCharacteristic Equation Roots<\/th>\r\n General Solution to the Differential Equation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n Distinct real roots,\u00a0[latex]\\lambda_1[\/latex] and\u00a0[latex]\\lambda_2[\/latex]<\/td>\r\n [latex]y(x)=c_1e^{\\lambda_1x}+c_2e^{\\lambda_2x}[\/latex]<\/td>\r\n<\/tr>\r\n \r\n A repeated real root, [latex]\\lambda[\/latex]<\/td>\r\n [latex]y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}[\/latex]<\/td>\r\n<\/tr>\r\n \r\n Complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex]<\/td>\r\n [latex]y(x)=e^{\\lambda x}(c_1\\cos\\beta x+c_2\\sin\\beta x)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n \t
\r\n \t
\r\n \t
Second-Order Equations with Constant Coefficients<\/h2>\n
characteristic equation<\/h3>\n
Distinct Real Roots<\/h3>\n
Single Repeated Real Root<\/h3>\n
Complex Conjugate Roots<\/h3>\n
Summary of Results<\/h3>\n
summary of characteristic equation cases<\/h3>\n
\n\n
\n \nCharacteristic Equation Roots<\/th>\n General Solution to the Differential Equation<\/th>\n<\/tr>\n<\/thead>\n \n Distinct real roots,\u00a0[latex]\\lambda_1[\/latex] and\u00a0[latex]\\lambda_2[\/latex]<\/td>\n [latex]y(x)=c_1e^{\\lambda_1x}+c_2e^{\\lambda_2x}[\/latex]<\/td>\n<\/tr>\n \n A repeated real root, [latex]\\lambda[\/latex]<\/td>\n [latex]y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}[\/latex]<\/td>\n<\/tr>\n \n Complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex]<\/td>\n [latex]y(x)=e^{\\lambda x}(c_1\\cos\\beta x+c_2\\sin\\beta x)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n<\/strong><\/p>\n\n
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