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- Linear homogeneous differential equations of 2nd order Step-By-Step
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How to use it?
- Differential equation:
- Equation y'+7*y=sin(x)
- Equation x^3*y''+x^2*y'=1
- Equation 2*x^3*y'=(2*x^2-y^2)*y
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Equation (x^2-y^2)y'=2xy
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Identical expressions
- sec(x)dx+csc^ two (x)dy= zero
- sec(x)dx plus csc squared (x)dy equally 0
- sec(x)dx plus csc to the power of two (x)dy equally zero
- sec(x)dx+csc2(x)dy=0
- secxdx+csc2xdy=0
- sec(x)dx+csc²(x)dy=0
- sec(x)dx+csc to the power of 2(x)dy=0
- secxdx+csc^2xdy=0
- sec(x)dx+csc^2(x)dy=O
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Similar expressions
- sec(x)dx-csc^2(x)dy=0
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Expressions with functions
- sec
- sec^2xsecy=y'(ctgx)(siny)
- sec^2xdy+cscydx=0
- secydx+e^(-x)=0
- sec^2(x)*ctg(y)dx+sec^2(y)*tg(x)dy=0
- sec^2(x)*tg(y)dy+sec^2(y)*tg(x)dx
- dx
- dx/dt=3x^2(1+y^2)
- dx/dt=1+t/t^2×x
- dx-√(a^2-x^2)dy=0
- dx/dt=3x-y
- dx*sqrt(y^2+4)-dy*y=dy*x^2*y
- cosec
- csc(x)y'+x(y^2-1)/y=0
- dy
- dy/dx=y/x+coty/x
- dy/y^4=x^5*dx
- dy/dx=3x⁴+x²+1
- dy/1-3x^2=dx/2y
- dy-cosxdx=0
Differential equation sec(x)dx+csc^2(x)dy=0
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The solution
You have entered
[src]
2 d
csc (x)*--(y(x)) + sec(x) = 0
dx
$$\csc^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + \sec{\left(x \right)} = 0$$
csc(x)^2*y' + sec(x) = 0
Detail solution
Divide both sides of the equation by the multiplier of the derivative of y':
$$\csc^{2}{\left(x \right)}$$
We get the equation:
y' = $$- \frac{\sec{\left(x \right)}}{\csc^{2}{\left(x \right)}}$$
This differential equation has the form:
It is solved by multiplying both sides of the equation by dx:
And by using the integrals of the both equation sides:
or
In this case,
f(x) = $$- \frac{\sec{\left(x \right)}}{\csc^{2}{\left(x \right)}}$$
Consequently, the solution will be
y = $$\int \left(- \frac{\sec{\left(x \right)}}{\csc^{2}{\left(x \right)}}\right)\, dx$$
Detailed solution of the integral
or
y = $$\frac{\log{\left(\sin{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sin{\left(x \right)} + 1 \right)}}{2} + \sin{\left(x \right)}$$ + C1
where C1 is constant, independent of x
$$\csc^{2}{\left(x \right)}$$
We get the equation:
y' = $$- \frac{\sec{\left(x \right)}}{\csc^{2}{\left(x \right)}}$$
This differential equation has the form:
y' = f(x)
It is solved by multiplying both sides of the equation by dx:
y'dx = f(x)dx, or
d(y) = f(x)dx
And by using the integrals of the both equation sides:
∫ d(y) = ∫ f(x) dx
or
y = ∫ f(x) dx
In this case,
f(x) = $$- \frac{\sec{\left(x \right)}}{\csc^{2}{\left(x \right)}}$$
Consequently, the solution will be
y = $$\int \left(- \frac{\sec{\left(x \right)}}{\csc^{2}{\left(x \right)}}\right)\, dx$$
Detailed solution of the integral
or
y = $$\frac{\log{\left(\sin{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sin{\left(x \right)} + 1 \right)}}{2} + \sin{\left(x \right)}$$ + C1
where C1 is constant, independent of x
The answer
[src]
log(-1 + sin(x)) log(1 + sin(x))
y(x) = C1 + ---------------- - --------------- + sin(x)
2 2
$$y{\left(x \right)} = C_{1} + \frac{\log{\left(\sin{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sin{\left(x \right)} + 1 \right)}}{2} + \sin{\left(x \right)}$$
The classification
factorable
nth algebraic
separable
1st exact
1st linear
1st power series
lie group
nth linear euler eq nonhomogeneous variation of parameters
nth algebraic Integral
separable Integral
1st exact Integral
1st linear Integral
nth linear euler eq nonhomogeneous variation of parameters Integral