# What is the general solution of the differential equation? # (4x+3y+15)dx + (2x + y +7)dy = 0 #

# 17/34 ln abs(((y+1)/(x+3))^2-(y+1)/(x+3)-4) - 5sqrt(17)/34 ln abs(2(y+1)/(x+3)+sqrt(17)-1) + 5sqrt(17)/34 ln abs(2 (y+1)/(x+3)-sqrt(17)-1) = ln abs(u) + C #

We have:

which we could type as:

It is not possible to use our standard toolkit for DEs, but we can use a transformation to take the constants out of the linear denominator and numerator.

Think about the simultaneous equations.

We thus carry out two linear transformations:

Furthermore, when we replace in the DE [A], we obtain

By inserting this replacement into our altered DE [B], we obtain:

Now that this is a separable DE, we can divide and rearrange the variables to obtain:

The RHGS integral is easily integrable after splitting into partial fractions (omitted), and the LHS is a standard integral. We discover that:

Next, going back to our previous substitution, we have:

Thus:

If you'd like, you can simplify this even more.

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To find the general solution of the given differential equation, you first need to check if it is exact. If it is exact, integrate with respect to one variable to find a function of two variables, and then solve for the constant of integration. If it is not exact, use an integrating factor to make it exact, then proceed with the same steps. Once you have found the general solution, express it in terms of y as a function of x, or x as a function of y.

For the given equation: ( (4x + 3y + 15)dx + (2x + y + 7)dy = 0 )

The equation is not exact. To make it exact, we need to find an integrating factor ( \mu(x, y) ). The integrating factor is given by:

[ \mu(x, y) = e^{\int \frac{M_y - N_x}{N} , dx} ]

where ( M ) and ( N ) are the coefficients of ( dx ) and ( dy ), respectively.

In this case, ( M = 4x + 3y + 15 ) and ( N = 2x + y + 7 ).

[ M_y = 3 \quad \text{and} \quad N_x = 2 ]

[ \mu(x, y) = e^{\int \frac{3 - 2}{2x + y + 7} , dx} = e^{\int \frac{1}{2x + y + 7} , dx} ]

Integrating ( \frac{1}{2x + y + 7} ) with respect to ( x ) gives ( \ln|2x + y + 7| ). Thus, the integrating factor is ( \mu(x, y) = |2x + y + 7| ).

Multiply the given differential equation by the integrating factor:

[ (4x + 3y + 15)|2x + y + 7|dx + (2x + y + 7)|2x + y + 7|dy = 0 ]

This equation is exact. Now, integrate with respect to ( x ) to find the general solution.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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