The general solution of the differential equation d y d x = x + y - 3 x + y - 7 is

2 log x + y - 5 = 3 x + 2 y + C

log x + y - 3 = 3 x + y - 2 2 + C

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A curve passes through the point $(1, \frac{π}{6}) .$ Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + s e c (\frac{y}{x}), x > 0$ Then, the equation of the curve is

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Important Questions on Differential Equations

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of the differential equation

\frac{d y}{d x} = \frac{2 x y - y^{2}}{2 x y - x^{2}},

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The equation of the curve which passes through point

(1, 0)

and has tangent with the slope

1 + \frac{y}{x} + {(\frac{y}{x})}^{2}

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The general solution of the differential equation

(1 + e^{\frac{x}{y}}) d x + (1 - \frac{x}{y}) e^{x / y} d y = 0

(C

is an arbitrary constant

)

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let

y = y (x)

be the solution of the differential equation

x \tan (\frac{y}{x}) d y = (y \tan (\frac{y}{x}) - x) d x

- 1 \leq x \leq 1, y (\frac{1}{2}) = \frac{π}{6} .

Then the area of the region bounded by the curves

x = 0, x = \frac{1}{\sqrt{2}}

and

y = y (x)

in the upper half plane is:

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of

\frac{d y}{d x} = \frac{y^{2}}{x y - x^{2}}

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Let

C_{1}

be the curve obtained by the solution of differential equation

2 x y \frac{d y}{d x} = y^{2} - x^{2}, x > 0

. Let the curve

C_{2}

be the solution of

\frac{2 x y}{x^{2} - y^{2}} = \frac{d y}{d x}

. If both the curves pass through

(1, 1),

then the area (in sq. units) enclosed by the curves

C_{1}

and

C_{2}

is equal to :

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

x \sin (\frac{y}{x}) d y = [y \sin (\frac{y}{x}) - x] d x, x > 0

and

y (1) = \frac{π}{2},

then the value of

\cos (\frac{y}{x})

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

Solution of the differential equation

x \frac{d y}{d x} = y - x \tan (\frac{y}{x})

will be

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The general solution of the differential equation

\frac{d y}{d x} = \frac{x + y - 3}{x + y - 7}

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The general solution of the differential equation

(x^{2} + x y) y^{'} = y^{2}

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

y \frac{d y}{d x} = x [\frac{y^{2}}{x^{2}} + \frac{ϕ (\frac{y^{2}}{x^{2}})}{ϕ^{'} (\frac{y^{2}}{x^{2}})}], x > 0, ϕ > 0,

and

y (1) = - 1,

then

ϕ (\frac{y^{2}}{4})

is equal to:

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

\frac{d y}{d x} = \frac{x y}{x^{2} + y^{2}}; y (1) = 1;

then a value of

x

satisfying

y (x) = e

is:

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of the differential equation

\frac{d y}{d x} = \frac{x y}{x^{2} + y^{2}}

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

\frac{d y}{d x} = \frac{y + x \tan \frac{y}{x}}{x}

, then

\sin \frac{y}{x} =

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of the differential equation

x \frac{d y}{d x} = y (\log_{e} y - \log_{e} x + 1)

will be

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The general solution of the differential equation

(x^{2} + y^{2}) d x - 2 x y d y = 0

MEDIUM

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

A solution of the differential equation

2 x y d y - (x^{2} + y^{2}) d x = 0

satisfying

y (1) = 1

is given by

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The solution of the differential equation

x^{2} \frac{d y}{d x} = y^{2} + x y

HARD

Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

The general solution of

\frac{dy}{dx} = \frac{2 x - y}{x + 2 y}

is given by

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Mathematics>Integral Calculus>Differential Equations>Methods of Solving First Order, First Degree Differential Equation

A curve passes through the point

(1, \frac{π}{6})

. Let the slope of the curve at each point

(x, y)

\frac{y}{x} + \sec (\frac{y}{x}), x > 0

. Then the equation of the curve is

A curve passes through the point 1,π6. Let the slope of the curve at each point (x, y) be yx+sec yx, x>0 Then, the equation of the curve is

Important Questions on Differential Equations

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A curve passes through the point $(1, \frac{π}{6}) .$ Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x} + s e c (\frac{y}{x}), x > 0$ Then, the equation of the curve is