Another approach would be to use an integrating factor (which incidentally is a method devised by Euler). The equation to solve is y' + y = x²eˣ Multiplying both sides by eˣ we have y'·eˣ + y·eˣ = x²·e²ˣ The left hand side is now the derivative of y·eˣ so integrating both sides we get y·eˣ = (½x² − ½x + ¼)·e²ˣ + c₁ and multiplying both sides by e⁻ˣ we have y = (½x² − ½x + ¼)·eˣ + c₁·e⁻ˣ and the differential equation is solved. Note that the general solution of the first order linear inhomogeneous differential equation only contains a _single_ constant. Your addition of a second constant at the end of your solution is wrong: clearly there can exist no two solutions of this differential equation which only differ by a constant. Alternatively, you could solve the homogeneous equation by means of its characteristic equation as you did in the video (again, a method devised by Euler) and then find a particular solution of the inhomogeneous equation by making an _Ansatz_ yₚ = (ax² + bx + c)·eˣ and substituting this in the inhomogeneous equation together with its derivative yₚ' = (ax² + (2a + b)x + (b + c))·eˣ Equating coefficients we then have 2a = 1 2a + 2b = 0 b + 2c = 0 which gives a = ½, b = −½, c = ¼, so we have a particular solution yₚ = (½x² − ½x + ¼)·eˣ and together with the general solution of the homogeneous equation yₕ = c₁·e⁻ˣ this again gives the general solution y = yₕ + yₚ of the inhomogeneous equation as y = c₁·e⁻ˣ + (½x² − ½x + ¼)·eˣ
Actually, the problem is that the constant should have been added at the end of v(x) which is then multiplied by e^(-x) in the particular solution. This would, then, make the final term of the general solution ce^(-x). And this will satisfy the original equation.
The product of a polynomial and e^x is a again a product of a polynomial and e^x, with the polynomial having the same degree. So I made the Ansatz y(x) = (ax^2 + bx + c) * e^x y'(x) = (ax^2 + bx + c) * e^x + (2ax + b) * e^x = (ax^2 + (2a + b)x + (b + c)) * e^x The sum is y'(x) + y(x) = (2ax^2 + (2a + 2b)x + (b + 2c)) * e^x And shall be equal to x^2 * e^x = (1*x^2 + 0*x + 0) * e^x Comparing the coefficients yields 2a = 1 a = 1/2 2a + 2b = 0 a + b = 0 1/2 + b = 0 b = -1/2 b + 2c = 0 -1/2 + 2c = 0 2c = 1/2 c = 1/4 Altogether, I get the solution function y(x) = (1/2*x^2 - 1/2*x + 1/4) * e^x = 1/4 * (2x^2 - 2x + 1) * e^x and y'(x) = 1/4 * [ (2x^2 - 2x + 1) * e^x + (4x - 2) * e^x ] = 1/4 * (2x^2 + 2x - 1) * e^x Their sum is indeed 1/4 * 4x^2 * e^x = x^2 * e^x
The +c constant in the final expression of the general solution must not be there. Indeed if you consider the linear operator L(y) = y’+y, L(c1 e^-x) = 0, L(1/2(x2-x+1/2)e^x) = x2e^x and L(c) = c. Therefore, given the linearity of L, and assuming the reported form of y_g(x), L(y_g(x)) = x2e^x +c, which means y_g would not satisfy the given equation.
Yes, the problem is he didn't add the constant term when he solved for v(x). Of course, that makes the final term of the general equation ce^(-x), which is extraneous, since it can just be added to the first term of the general solution to form a new arbitrary constant.
First order linear equation We can use : undetermined coeffciens (Homogegeous part has constant coefficients and non-homogeneous part is in the form P(x)e^{λx} , where λ \in C and P(x) is polynomial, because of superposition we can also have linear combination of functions in mentioned form) We can use integrating factor or variation of parameter For variation of parameter we need to solve homogeneous equation first
You could have used the y particular method in the first example you just have to make sure that it is linearly independent from y homogeneous. And in this case it is. Even if it’s not you just need to multiply by x until it becomes linearly independent.
You sure that the "+ C" belongs? We have a "+ C" on normal integrals because it's the homogeneous solution to equations of the form "y' = f(x)". But this equation isn't in that form, and the homogeneous solution is something else.
The simple substitution y(x)=z(x)•exp(x) converts the original DE into z'+2•z=x² What could be easily solved The particular solution can be found in form z(x) = x²+a•x+b z'(x) = 2x+a what implies x²+a•x+b+4x+2a=x² a=-4 b=8 z(x)=x²-4x+8 The general solution is z(x)= C•exp(-2x)+x²-4x+8 y(x)= C•exp(-x)+(x²-4x+8)•exp(x) Where the constant C must be found from the initial condition: y(0)=C+8
Another approach would be to use an integrating factor (which incidentally is a method devised by Euler). The equation to solve is
y' + y = x²eˣ
Multiplying both sides by eˣ we have
y'·eˣ + y·eˣ = x²·e²ˣ
The left hand side is now the derivative of y·eˣ so integrating both sides we get
y·eˣ = (½x² − ½x + ¼)·e²ˣ + c₁
and multiplying both sides by e⁻ˣ we have
y = (½x² − ½x + ¼)·eˣ + c₁·e⁻ˣ
and the differential equation is solved. Note that the general solution of the first order linear inhomogeneous differential equation only contains a _single_ constant. Your addition of a second constant at the end of your solution is wrong: clearly there can exist no two solutions of this differential equation which only differ by a constant.
Alternatively, you could solve the homogeneous equation by means of its characteristic equation as you did in the video (again, a method devised by Euler) and then find a particular solution of the inhomogeneous equation by making an _Ansatz_
yₚ = (ax² + bx + c)·eˣ
and substituting this in the inhomogeneous equation together with its derivative
yₚ' = (ax² + (2a + b)x + (b + c))·eˣ
Equating coefficients we then have
2a = 1
2a + 2b = 0
b + 2c = 0
which gives a = ½, b = −½, c = ¼, so we have a particular solution
yₚ = (½x² − ½x + ¼)·eˣ
and together with the general solution of the homogeneous equation
yₕ = c₁·e⁻ˣ
this again gives the general solution y = yₕ + yₚ of the inhomogeneous equation as
y = c₁·e⁻ˣ + (½x² − ½x + ¼)·eˣ
very good!
you can multiply both sides by e^x and then the left side becomes the derivative of y*e^x which makes it easy to solve
Great! You use an IF
You have one extra constant C on the last line which screws up the solution.
Actually, the problem is that the constant should have been added at the end of v(x) which is then multiplied by e^(-x) in the particular solution. This would, then, make the final term of the general solution ce^(-x). And this will satisfy the original equation.
Actually, I just realized that final term (ce^(-x)) is extraneous, since it can just be added to the first term.
The product of a polynomial and e^x is a again a product of a polynomial and e^x, with the polynomial having the same degree.
So I made the Ansatz
y(x) = (ax^2 + bx + c) * e^x
y'(x)
= (ax^2 + bx + c) * e^x + (2ax + b) * e^x
= (ax^2 + (2a + b)x + (b + c)) * e^x
The sum is
y'(x) + y(x)
= (2ax^2 + (2a + 2b)x + (b + 2c)) * e^x
And shall be equal to
x^2 * e^x = (1*x^2 + 0*x + 0) * e^x
Comparing the coefficients yields
2a = 1
a = 1/2
2a + 2b = 0
a + b = 0
1/2 + b = 0
b = -1/2
b + 2c = 0
-1/2 + 2c = 0
2c = 1/2
c = 1/4
Altogether, I get the solution function
y(x)
= (1/2*x^2 - 1/2*x + 1/4) * e^x
= 1/4 * (2x^2 - 2x + 1) * e^x
and
y'(x)
= 1/4 * [ (2x^2 - 2x + 1) * e^x + (4x - 2) * e^x ]
= 1/4 * (2x^2 + 2x - 1) * e^x
Their sum is indeed
1/4 * 4x^2 * e^x
= x^2 * e^x
Same!
The +c constant in the final expression of the general solution must not be there.
Indeed if you consider the linear operator L(y) = y’+y, L(c1 e^-x) = 0, L(1/2(x2-x+1/2)e^x) = x2e^x and L(c) = c. Therefore, given the linearity of L, and assuming the reported form of y_g(x), L(y_g(x)) = x2e^x +c, which means y_g would not satisfy the given equation.
Yes, the problem is he didn't add the constant term when he solved for v(x). Of course, that makes the final term of the general equation ce^(-x), which is extraneous, since it can just be added to the first term of the general solution to form a new arbitrary constant.
First order linear equation
We can use :
undetermined coeffciens
(Homogegeous part has constant coefficients and non-homogeneous part is in the form P(x)e^{λx} , where λ \in C and P(x) is polynomial,
because of superposition we can also have linear combination of functions in mentioned form)
We can use integrating factor or variation of parameter
For variation of parameter we need to solve homogeneous equation first
You could have used the y particular method in the first example you just have to make sure that it is linearly independent from y homogeneous. And in this case it is. Even if it’s not you just need to multiply by x until it becomes linearly independent.
Nice!
You sure that the "+ C" belongs? We have a "+ C" on normal integrals because it's the homogeneous solution to equations of the form "y' = f(x)". But this equation isn't in that form, and the homogeneous solution is something else.
The simple substitution
y(x)=z(x)•exp(x)
converts the original DE into
z'+2•z=x²
What could be easily solved
The particular solution can be found in form
z(x) = x²+a•x+b
z'(x) = 2x+a
what implies
x²+a•x+b+4x+2a=x²
a=-4
b=8
z(x)=x²-4x+8
The general solution is
z(x)= C•exp(-2x)+x²-4x+8
y(x)= C•exp(-x)+(x²-4x+8)•exp(x)
Where the constant C must be found from the initial condition:
y(0)=C+8
λ=-1..yo(x)=ce^(-x)...siccome -1 è diverso dall esponente di e^x,risulta semplicemente yp(x)=(ax^2+bx+c)e^x=(1/2x^2-1/2x+1/4)e^x
👌
Linear DE
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