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    Let L be defined by L(y)=y\prime\prime+ay\prime+by for a,b \in \mathbb{R} constants.

    Let R be a complex valued function so that R(x)=P(x)+iQ(x) for functions P(x) and Q(x). Prove the function f(x)=u(x)+iv(x) satisfies the differential equation L(y)=R(x) \enspace x \in I if and only if u and v satisfy the differential equations L(u)=P(x) \enspace L(v)=Q(x) \enspace x \in I.

    I literally don't even know where to start, are there some formula or trick that can help?
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    (Original post by AishaGirl)
    Let R be a complex valued function so that R(x)+iQ(x) for functions P(x) and Q(x).
    I may be wrong or misreading something here, but should R(x)+iQ(x) read R(x) = P(x)+iQ(x)?
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    (Original post by crashMATHS)
    I may be wrong or misreading something here, but should R(x)+iQ(x) read R(x) = P(x)+iQ(x)?
    Yes you are right, sorry it was a typo I fixed it now.
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    (Original post by AishaGirl)
    Let L be defined by L(y)=y\prime\prime+ay\prime+by for a,b \in \mathbb{R} constants.

    Let R be a complex valued function so that R(x)=P(x)+iQ(x) for functions P(x) and Q(x). Prove the function f(x)=u(x)+iv(x) satisfies the differential equation L(y)=R(x) \enspace x \in I if and only if u and v satisfy the differential equations L(u)=P(x) \enspace L(v)=Q(x) \enspace x \in I.

    I literally don't even know where to start, are there some formula or trick that can help?
    Hint: a+ib=c+id if and only if a=c,b=d
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    ok so I think I have an idea how to start.

    If I assume that y=f(x)=u(x)+iv(x) satisfies the differential equation then can I say something like

    L(y)=R(x) \implies f\prime\prime(x)+af\prime(x)+bf(  x)=P(x)+iQ(x)

    \implies u\prime\prime(x)+iv\prime\prime(  x)+au \prime(x)+iav\prime(x)+bu\prime(  x)+bv\prime(x)=P(x)+iQ(x)

    Is the correct way to go?
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    (Original post by AishaGirl)
    Is the correct way to go?
    Yes. Now collect real and imaginary parts.
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    (Original post by atsruser)
    Yes. Now collect real and imaginary parts.
    Like this?

    u\prime\prime(x)+au\prime(x)+bu(  x)=P(x)

    v\prime\prime(x)+av\prime(x)+bv(  x)=Q(x)

    So L(u)=P(x) and L(v)=Q(x) ?

    Looks good so far?
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    I think I have proved it now.

    Assume L(u)=P(x) and L(v)=Q(x) for all x\in I then

    u\prime\prime(x)+au \prime(x)+bu(x)+i(v\prime\prime(  x)+av \prime(x)+bv(x))=P(x)+iQ(x)

     \implies f\prime\prime(x)+af\prime(x)+bf(  x)=R(x) \implies L(y)=R(x) for all x\in I

    Is this correct? It's quite the mouthful

    atsruser alow RDKGames DFranklin can someone tell me if it's correct?
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    (Original post by AishaGirl)
    I think I have proved it now.

    Assume L(u)=P(x) and L(v)=Q(x) for all x\in I then

    u\prime\prime(x)+au \prime(x)+bu(x)+i(v\prime\prime(  x)+av \prime(x)+bv(x))=P(x)+iQ(x)

     \implies f\prime\prime(x)+af\prime(x)+bf(  x)=R(x) \implies L(y)=R(x) for all x\in I

    Is this correct? It's quite the mouthful

    atsruser alow RDKGames DFranklin can someone tell me if it's correct?
    Looks good
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    (Original post by crashMATHS)
    Looks good
    Cheers!
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    (Original post by AishaGirl)
    I think I have proved it now.

    Assume L(u)=P(x) and L(v)=Q(x) for all x\in I then

    u\prime\prime(x)+au \prime(x)+bu(x)+i(v\prime\prime(  x)+av \prime(x)+bv(x))=P(x)+iQ(x)

     \implies f\prime\prime(x)+af\prime(x)+bf(  x)=R(x) \implies L(y)=R(x) for all x\in I
    You're halfway there. The question is an "if and only if" proof. Suppose A and B are statements, and A is true if and only if B is true - that means that A implies B and that B implies A i.e.

    A \Rightarrow B and B \Rightarrow A or equivalently A \Leftrightarrow B

    You have proved the "if" part i.e. A \Rightarrow B part. You still need to prove the "only if" part i.e. B \Rightarrow A. To do so, you need to show that if f(x) satisfies the given DE, then L(u) = P(x), L(v)=Q(x)
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    (Original post by atsruser)
    You're halfway there. The question is an "if and only if" proof. Suppose A and B are statements, and A is true if and only if B is true - that means that A implies B and that B implies A i.e.

    A \Rightarrow B and B \Rightarrow A or equivalently A \Leftrightarrow B

    You have proved the "if" part i.e. A \Rightarrow B part. You still need to prove the "only if" part i.e. B \Rightarrow A. To do so, you need to show that if f(x) satisfies the given DE, then L(u) = P(x), L(v)=Q(x)
    I believe she's done both directions on two separate posts
 
 
 
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