How to solve this diffrential equation using parts formula?dydx+y20=50(1+cosx)

Step 1Hint:∫eaxcos⁡(bx)&nbsp;dx&nbsp;=deaxa2+b2[acos⁡(bx)+bsin⁡(bx)]+cProof:Let,I=∫eax⏟ucos⁡(bx)&nbsp;dx&nbsp;⏟dv=uv−∫vdu⇒I=eax[d1bsin⁡(bx)]−dab∫eaxsin⁡(bx)&nbsp;dx&nbsp;=eax[d1bsin⁡(bx)]−(dabeax[−d1bcos⁡(bx)]−da2b2∫eaxsin⁡(bx)&nbsp;dx&nbsp;)⇒I=eax(acos(bx)+bsin(bx)b2+a2b2I⇒I(1+a2b2)=eax(acos(bx)+bsin(bx)b2⇒I=eaxa2+b2acos(bx)+bsin(bx)

"How to solve this diffrential equation using parts..." solved and explained

Octavio Chen

Answered question

2022-03-29

How to solve this diffrential equation using parts formula?
$\frac{d y}{d x} + \frac{y}{20} = 50 (1 + \cos x)$

Answer & Explanation

Deon Frost

Beginner2022-03-30Added 6 answers

Step 1
Hint:
$\int e^{a x} \cos (b x) d x = d \frac{e^{a x}}{a^{2} + b^{2}} [a \cos (b x) + b \sin (b x)] + c$
Proof:
Let,
$I = \int \underset{u}{\underset{⏟}{e^{a x}}} \underset{d v}{\underset{⏟}{\cos (b x) d x}} = u v - \int v d u$
$\Rightarrow I = e^{a x} [d \frac{1}{b} \sin (b x)] - d \frac{a}{b} \int e^{a x} \sin (b x) d x$
$= e^{a x} [d \frac{1}{b} \sin (b x)] - (d \frac{a}{b} e^{a x} [- d \frac{1}{b} \cos (b x)] - d \frac{a^{2}}{b^{2}} \int e^{a x} \sin (b x) d x)$
$\Rightarrow I = \frac{e^{a x} (a \cos (b x) + b \sin (b x)}{b^{2}} + \frac{a^{2}}{b^{2}} I$
$\Rightarrow I (1 + \frac{a^{2}}{b^{2}}) = \frac{e^{a x} (a \cos (b x) + b \sin (b x)}{b^{2}}$
$\Rightarrow I = \frac{e^{a x}}{a^{2} + b^{2}} [a \cos (b x) + b \sin (b x)]$

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