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Grade 12th passDifferential Calculus

Solve the differential equation dy/dx=sin(x+y)
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Profile image of Shivangi singh
7 Years agoGrade 12th pass
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2 Answers

Profile image of Khimraj
ApprovedApproved Tutor Answer7 Years ago
now assume tanv/2 = t and solve further.   This is the solution.\int \frac{sec^2\frac{v}{2}dv}{1+2tan\frac{v}{2}+tan^{2}\frac{v}{2}}= \int dx \int \frac{dv}{1+\frac{2tan\frac{v}{2}}{1+tan^2\frac{v}{2}}}= \int dx \int \frac{dv}{1+sinv}= \int dx Integrate both sides\frac{dv}{1+sinv}= dx Using variable and separable method\frac{dv}{dx}-1= sin(v) \frac{dy}{dx} = sin(x+y)Put x+y = v1+dy/dx = dv/dxdy/dx = dv/dx – 1
Profile image of Khimraj
7 Years ago
Put x+y = v
1+dy/dx = dv/dx
dy/dx = dv/dx – 1
 \frac{dy}{dx} = sin(x+y)
\frac{dv}{dx}-1= sin(v)
Using variable and separable method
\frac{dv}{1+sinv}= dx 
Integrate both sides
 \int \frac{dv}{1+sinv}= \int dx
\int \frac{dv}{1+\frac{2tan\frac{v}{2}}{1+tan^2\frac{v}{2}}}= \int dx 
\int \frac{sec^2\frac{v}{2}dv}{1+2tan\frac{v}{2}+tan^{2}\frac{v}{2}}= \int dx
now assume tanv/2 = t and solve further.   This is the solution.