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Question Number 154785 by SANOGO last updated on 21/Sep/21

soit: y′+tan(x)y=sin(2x) ,avec f(0)=1 alors f(π)=?

soit:y′+tan(x)y=sin(2x),avecf(0)=1alorsf(π)=?

Commented by SANOGO last updated on 21/Sep/21

merci bien le dur

mercibienledur

Commented by tabata last updated on 21/Sep/21

you are welcome

youarewelcome

Commented by tabata last updated on 21/Sep/21

p(x) = tan(x) , Q(x) = sin(2x) (I.f) = e^(∫ p(x)dx) = e^(∫ tan(x)) = e^(−ln∣cos(x)∣) = sec(x) y = ((∫ (I.f) Q (x) dx)/((I.f))) = ((∫ 2 sin (x) dx )/(sec(x))) = − 2 cos^2 (x) + c cos(x) ∴ y = − 2 cos^2 (x) + c cos(x) f(0) = 1 ⇒ 1 = −2 + c ⇒ c = 3 ∴ y = − 2 cos^2 (x) + 3 cos(x) f ( 𝛑 ) = − 2 ( cos(𝛑))^2 + 3 cos(𝛑) f ( 𝛑 ) = − 2 − 3 = −5 ⟨ M . T ⟩

p(x)=tan(x),Q(x)=sin(2x)(I.f)=e∫p(x)dx=e∫tan(x)=e−ln∣cos(x)∣=sec(x)y=∫(I.f)Q(x)dx(I.f)=∫2sin(x)dxsec(x)=−2cos2(x)+ccos(x)∴y=−2cos2(x)+ccos(x)f(0)=1⇒1=−2+c⇒c=3∴y=−2cos2(x)+3cos(x)f(π)=−2(cos(π))2+3cos(π)f(π)=−2−3=−5⟨M.T⟩

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