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Solve the following differential equatio...

Solve the following differential equation: `(dy)/(dx)=(cos^2x-sin^2x)cos^2y`

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To solve the differential equation \(\frac{dy}{dx} = (\cos^2 x - \sin^2 x) \cos^2 y\), we will use the method of separation of variables. Here are the steps: ### Step 1: Separate the variables We start by rewriting the equation to separate the variables \(y\) and \(x\): \[ \frac{dy}{\cos^2 y} = (\cos^2 x - \sin^2 x) dx \] ...
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RD SHARMA-DIFFERENTIAL EQUATION-Solved Examples And Exercises
  1. Solve the following differential equation: dy+(x+1)(y+1)dx=0

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  2. Solve the following differential equation: (x-1)(dy)/(dx)=2x^3y

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  3. Solve the following differential equation: (dy)/(dx)=(cos^2x-sin^2x)co...

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  4. Solve the following differential equation: cosec x log y(dy)/(dx)+\ x...

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  5. Solve the following differential equation: y(1-x^2)(dy)/(dx)=x(1+y^2)

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  6. Solve the differential equation y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)

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  7. Solve the following differential equation: (1+y^2)tan^(-1)dx+2y(1+x^2)...

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  8. Solve the following initial value problem: (dy)/(dx)=y \ tan2x ,\ y(0)...

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  9. Solve the following initial value problem: x y(dy)/(dx)=y+2,\ y(2)=0

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  10. Solve the following initial value problem: (d r)/(dt)=\ -r t ,\ r(0) =...

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  11. Solve the following initial value problem: (dy)/(dx)=ytanx ,\ y(0)=1

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  12. Solve the following initial value problem: (dy)/(dx)=2e^(2x)y^2,\ y(0)...

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  13. Solve the following initial value problem: (dy)/(dx)=2x y ,\ y(0)=1

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  14. Solve the following initial value problem: (dy)/(dx)=1+x^2+y^2+x^2y^2,...

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  15. Solve the following initial value problem: x y(dy)/(dx)=x^2+2y^2),\ y(...

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  16. Solve the following initial value problem: (dy)/(dx)=1+x+y^2+xy^2 when...

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  17. Solve the following initial value problem: 2(y+3)-x y(dy)/(dx)=0,\ y(1...

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  18. Solve the following initial value problem: 2x(dy)/(dx)=3y ,\ y(1)=2

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  19. Solve the following initial value problem: (dy)/(dx)=2e^x y^3,\ y(0)=1...

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  20. Solve the following initial value problem: (dy)/(dx)=ysin2x ,\ y(0)=1

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