Answer:
[tex]\qquad\boxed{ \sf{ \: \bf \: y \: = \dfrac{1}{2}tan\dfrac{\theta}{2} + \dfrac{1}{2} \: cot\dfrac{\theta}{2} \: \: }}\\ \\ [/tex]
Step-by-step explanation:
Given differential equation is
[tex]\sf \: sin\theta \dfrac{dy}{d\theta} + y = tan\dfrac{1}{2}\theta \\ \\ [/tex]
Divide whole equation by sin[tex]\theta [/tex], we get
[tex]\sf \: \dfrac{dy}{d\theta} +(cosec\theta ) y =(cosec\theta ) tan\dfrac{1}{2}\theta \\ \\ [/tex]
So, its a linear differential equation. On comparing with
[tex]\sf \: \dfrac{dy}{d\theta} +py = q, \: we \: get \\ \\ [/tex]
[tex]\sf \: p = cosec\theta \\ \\ [/tex]
[tex]\sf \: q = tan\dfrac{\theta }{2} \\ \\ [/tex]
So, Integrating Factor is
[tex]\sf \: IF = {e}^{ \: \displaystyle\int\sf p \: d\theta } \\ \\ [/tex]
[tex]\sf \: IF = {e}^{ \: \displaystyle\int\sf cosec\theta \: d\theta } \\ \\ [/tex]
[tex]\sf \: IF = {e}^{ \: ln \: tan\dfrac{\theta }{2} } \\ \\ [/tex]
[tex]\sf\implies \: \sf \: IF = tan\dfrac{\theta }{2} \\ \\ [/tex]
Thus, Solution of differential equation is
[tex]\sf \: y \times IF = \displaystyle\int\sf q \times IF \: d\theta \\ \\ [/tex]
[tex]\sf \: y \times tan\dfrac{\theta}{2} = \displaystyle\int\sf cosec\theta \times tan\dfrac{\theta}{2}\times tan\dfrac{\theta}{2} \: d\theta \\ \\ [/tex]
[tex]\sf \: y \times tan\dfrac{\theta}{2} = \displaystyle\int\sf \frac{1}{sin\theta } \times tan^{2} \dfrac{\theta}{2}\: d\theta \\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \displaystyle\int\sf \frac{1 + {tan}^{2}\dfrac{\theta}{2} }{2tan\dfrac{\theta}{2} } \times tan^{2} \dfrac{\theta}{2}\: d\theta \\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \frac{1}{2} \displaystyle\int\sf {sec}^{2}\dfrac{\theta}{2} \times tan^{} \dfrac{\theta}{2}\: d\theta \\ \\ [/tex]
Now, to evaluate this integral, Substitute
[tex]\sf \: tan\dfrac{\theta}{2} = z \\ \\ [/tex]
[tex]\sf\implies \sf \: {sec}^{2} \dfrac{\theta}{2} \times \dfrac{1}{2} \: d\theta = dz \\ \\ [/tex]
So, above integral can be rewritten as
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \displaystyle\int\sf z \: dz \\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \dfrac{ {z}^{2} }{2} + c\\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \dfrac{ {tan}^{2}\dfrac{\theta}{2} }{2} + c\\ \\ [/tex]
[tex]\sf\implies \sf \: y \: = \dfrac{1}{2}tan\dfrac{\theta}{2} + c \: cot\dfrac{\theta}{2} - - - (1)\\ \\ [/tex]
On substituting y = 1 and [tex]\theta = \frac{\pie}{2} [/tex], we get
[tex]\sf\implies \sf \: 1 \: = \dfrac{1}{2}tan\dfrac{\pi}{4} + c \: cot\dfrac{\pi}{4}\\ \\ [/tex]
[tex] \sf \: 1 \: = \dfrac{1}{2} + c \: \\ \\ [/tex]
[tex] \sf \: 1 - \dfrac{1}{2} = c \: \\ \\ [/tex]
[tex]\sf\implies \: \sf \:c = \dfrac{1}{2}\: \\ \\ [/tex]
So, equation (1) can be rewritten as
[tex]\sf\implies \sf \: y \: = \dfrac{1}{2}tan\dfrac{\theta}{2} + \dfrac{1}{2} \: cot\dfrac{\theta}{2} \\ \\ [/tex]
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Verified answer
Answer:
[tex]\qquad\boxed{ \sf{ \: \bf \: y \: = \dfrac{1}{2}tan\dfrac{\theta}{2} + \dfrac{1}{2} \: cot\dfrac{\theta}{2} \: \: }}\\ \\ [/tex]
Step-by-step explanation:
Given differential equation is
[tex]\sf \: sin\theta \dfrac{dy}{d\theta} + y = tan\dfrac{1}{2}\theta \\ \\ [/tex]
Divide whole equation by sin[tex]\theta [/tex], we get
[tex]\sf \: \dfrac{dy}{d\theta} +(cosec\theta ) y =(cosec\theta ) tan\dfrac{1}{2}\theta \\ \\ [/tex]
So, its a linear differential equation. On comparing with
[tex]\sf \: \dfrac{dy}{d\theta} +py = q, \: we \: get \\ \\ [/tex]
[tex]\sf \: p = cosec\theta \\ \\ [/tex]
[tex]\sf \: q = tan\dfrac{\theta }{2} \\ \\ [/tex]
So, Integrating Factor is
[tex]\sf \: IF = {e}^{ \: \displaystyle\int\sf p \: d\theta } \\ \\ [/tex]
[tex]\sf \: IF = {e}^{ \: \displaystyle\int\sf cosec\theta \: d\theta } \\ \\ [/tex]
[tex]\sf \: IF = {e}^{ \: ln \: tan\dfrac{\theta }{2} } \\ \\ [/tex]
[tex]\sf\implies \: \sf \: IF = tan\dfrac{\theta }{2} \\ \\ [/tex]
Thus, Solution of differential equation is
[tex]\sf \: y \times IF = \displaystyle\int\sf q \times IF \: d\theta \\ \\ [/tex]
[tex]\sf \: y \times tan\dfrac{\theta}{2} = \displaystyle\int\sf cosec\theta \times tan\dfrac{\theta}{2}\times tan\dfrac{\theta}{2} \: d\theta \\ \\ [/tex]
[tex]\sf \: y \times tan\dfrac{\theta}{2} = \displaystyle\int\sf \frac{1}{sin\theta } \times tan^{2} \dfrac{\theta}{2}\: d\theta \\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \displaystyle\int\sf \frac{1 + {tan}^{2}\dfrac{\theta}{2} }{2tan\dfrac{\theta}{2} } \times tan^{2} \dfrac{\theta}{2}\: d\theta \\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \frac{1}{2} \displaystyle\int\sf {sec}^{2}\dfrac{\theta}{2} \times tan^{} \dfrac{\theta}{2}\: d\theta \\ \\ [/tex]
Now, to evaluate this integral, Substitute
[tex]\sf \: tan\dfrac{\theta}{2} = z \\ \\ [/tex]
[tex]\sf\implies \sf \: {sec}^{2} \dfrac{\theta}{2} \times \dfrac{1}{2} \: d\theta = dz \\ \\ [/tex]
So, above integral can be rewritten as
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \displaystyle\int\sf z \: dz \\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \dfrac{ {z}^{2} }{2} + c\\ \\ [/tex]
[tex]\sf \: y \: tan\dfrac{\theta}{2} = \dfrac{ {tan}^{2}\dfrac{\theta}{2} }{2} + c\\ \\ [/tex]
[tex]\sf\implies \sf \: y \: = \dfrac{1}{2}tan\dfrac{\theta}{2} + c \: cot\dfrac{\theta}{2} - - - (1)\\ \\ [/tex]
On substituting y = 1 and [tex]\theta = \frac{\pie}{2} [/tex], we get
[tex]\sf\implies \sf \: 1 \: = \dfrac{1}{2}tan\dfrac{\pi}{4} + c \: cot\dfrac{\pi}{4}\\ \\ [/tex]
[tex] \sf \: 1 \: = \dfrac{1}{2} + c \: \\ \\ [/tex]
[tex] \sf \: 1 - \dfrac{1}{2} = c \: \\ \\ [/tex]
[tex]\sf\implies \: \sf \:c = \dfrac{1}{2}\: \\ \\ [/tex]
So, equation (1) can be rewritten as
[tex]\sf\implies \sf \: y \: = \dfrac{1}{2}tan\dfrac{\theta}{2} + \dfrac{1}{2} \: cot\dfrac{\theta}{2} \\ \\ [/tex]