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Thread: Differential Equations: Solve dy/dx = e^(2x) - 3y and y=1 when x=0.

  1. #1

    Differential Equations: Solve dy/dx = e^(2x) - 3y and y=1 when x=0.

    Solve the following differential equation.
    dy/dx = e^(2x) - 3y and y=1 when x=0.
    Hint: Recognize this as a first-order linear differential equation and follow the general method for solving these and use the initial conditions to find the integration constant.

    Solution:
    y=
    (e^(2x))/5 + 4/(5e^(3x))

    Can someone please solve this step by step and explain how to get to the solution?

  2. #2
    Elite Member
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    Quote Originally Posted by MBhatti View Post
    Solve the following differential equation.
    dy/dx = e^(2x) - 3y and y=1 when x=0.
    Hint: Recognize this as a first-order linear differential equation and follow the general method for solving these and use the initial conditions to find the integration constant.

    Solution:
    y=
    (e^(2x))/5 + 4/(5e^(3x))

    Can someone please solve this step by step and explain how to get to the solution?
    Can you solve the homogeneous part, i.e.:

    y' + 3y = 0

    Please share your solution with us.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Do you not have a teacher or at least a text book? There is a pretty standard formula for converting any linear first order differential equation to a single derivative using an "integrating factor". Given [tex]\frac{dy}{dx}+ f(x)y= g(x)[/tex], multiply the equation by the "integrating factor" [tex]\mu(x)[/tex]: [tex]\mu \frac{dy}{dx}+ \mu f y= \mu g[/tex].

    A single derivative would be of the form [tex]\frac{d(\mu y)}{dx}[/tex] which, by the product rule, is equal to [tex]\mu \frac{dy}{dx}+ \frac{d\mu}{dx}y[/tex]. Comparing that to [tex]\mu \frac{dy}{dx}+ \mu f y[/tex], we see that we must have [tex]\frac{d\mu}{dx}= \mu f[/tex]. Solve that differential equation (which is a separable equation) for [tex]\mu[/tex].

    In fact, for any f(x) we can write that as [tex]\frac{d\mu}{\mu}= f(x)dx[/tex] and, integrating [tex]ln(\mu)= \int f(x)dx+ C[/tex]. Taking the exponential, [tex]\mu(x)= C'e^{\int f(x)dx}[/tex]. Since we only need an integrating factor, we can take C' to be 1.

    With this linear first order differential equation, [tex]\frac{dy}{dx}+ 3y= e^{2x}[/tex], f(x) is the constant 3 so the integrating factor is [tex]\mu(x)= e^{\int 3dx}= e^{3x}[/tex].

    Multiplying on both sides of the equation by that [tex]e^{3x}\frac{dy}{dx}+ 3e^{3x}y= \frac{d(e^{3x}y)}{dx}= e^{5x}[/tex]. Integrate [tex]d(e^{3x}y)= e^{5x}dx[/tex].

    Subhotosh Kahn is suggesting a different method that applies to linear equations of any order, 1 or higher. A linear equation of order "n" would be an equation where the function, y, and its derivatives up to nth derivative are multiplied by functions of x and added. It is "homogeneous" if that is equal to 0, "non-homogeneous" if not. The idea is that we can find the general solution to a non-homogeneous equation by finding the general solution to the corresponding homogeneous equation (drop the function of x that is not multiplied by y or a derivative of y) then add to that any one solution to the entire equation.

    Here the differential equation is [tex]\frac{dy}{dx}+ 3y= e^{2x}[/tex] is "non-homogeneous" because of the "[tex]e^{2x}[/tex]". Dropping that gives the "homogeneous" equation [tex]\frac{dy}{dx}+ 3y= 0[/tex]. That is the same as [tex]\frac{dy}{dx}= -3y[/tex]. We can separate that as [tex]\frac{dy}{y}= -3dx[/tex] and integrate. To find a single solution to the entire equation, seeing that the "non-homogeneous part" is [tex]e^{2x}[/tex], I would recommend setting [tex]y= Ae^{2x}[/tex], for A a constant, putting it into the equation, and seeing what A must be in order to satisfy the equation.

  4. #4
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    Quote Originally Posted by MBhatti View Post
    Can someone please solve this step by step and explain how to get to the solution?
    Please read the forum rules before asking such requests.
    A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin

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