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    The variables x and y satisfy the differential equation
    dy/dx= e^2x+y
    ,
    and y = 0 when x = 0. Solve the differential equation, obtaining an expression for y in terms of x.
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    (Original post by Reneewinters)
    The variables x and y satisfy the differential equation
    dy/dx= e^2x+y
    ,
    and y = 0 when x = 0. Solve the differential equation, obtaining an expression for y in terms of x.
    Bring the y to the other side. You have a linear equation and need to multiply through by an integrating factor.
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    (Original post by Reneewinters)
    The variables x and y satisfy the differential equation
    dy/dx= e^2x+y
    ,
    and y = 0 when x = 0. Solve the differential equation, obtaining an expression for y in terms of x.
    I'm having problems with uploading the image that contains the worked solution to this problem. So i will try my best to type up the solution instead.

    dy/dx = e^2x+y = (e^2x)(e^y) <--- laws of indices

    int (1/e^y)dy = int (e^2x) dx
    => int (e^-y) dy = int (e^2x) dx <--- "int" means integrate. Gather up y's on RHS and x's on LHS

    -e^-y = 0.5e^2x + C

    -e^0 = 0.5e^0 + C
    -1 = 0.5 + C
    C = -3/2 <--- use y=0 and x=0 to find value of C

    -e^-y = 0.5e^2x - 3/2
    e^-y = -0.5e^2x + 3/2
    -y = ln (-0.5e^2x + 3/2)
    y = -ln (-0.5e^2x + 3/2)
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    (Original post by dip0)
    I'm having problems with uploading the image that contains the worked solution to this problem. So i will try my best to type up the solution instead.

    dy/dx = e^2x+y = (e^2x)(e^y) <--- laws of indices

    int (1/e^y)dy = int (e^2x) dx
    => int (e^-y) dy = int (e^2x) dx <--- "int" means integrate. Gather up y's on RHS and x's on LHS

    -e^-y = 0.5e^2x + C

    -e^0 = 0.5e^0 + C
    -1 = 0.5 + C
    C = -3/2 <--- use y=0 and x=0 to find value of C

    -e^-y = 0.5e^2x - 3/2
    e^-y = -0.5e^2x + 3/2
    -y = ln (-0.5e^2x + 3/2)
    y = -ln (-0.5e^2x + 3/2)
    Thank you so much. This helps a lot.
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    (Original post by Reneewinters)
    Thank you so much. This helps a lot.
    You do realise that tiny hobbit's and dip0's posts cover two separate problems, right?

    Tiny hobbit's covers y'=e^{2x}+y

    dip0's covers y'=e^{2x+y}


    So really, please aim to add more brackets in the appropriate places when writing an expression like that, otherwise it is ambiguous as it stands and the responses you get may not correspond to the problem you actually have.
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    (Original post by RDKGames)
    You do realise that tiny hobbit's and dip0's posts cover two separate problems, right?

    Tiny hobbit's covers y'=e^{2x}+y

    dip0's covers y'=e^{2x+y}


    So really, please aim to add more brackets in the appropriate places when writing an expression like that, otherwise it is ambiguous as it stands and the responses you get may not correspond to the problem you actually have.
    I do realise that and hence I acknowledged dip0's response to my question. But I will be sure to keep in mind to add brackets more appropriately so there is no question of ambiguity left.
    Thank you for the advice.
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    (Original post by Reneewinters)
    Thank you so much. This helps a lot.
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