Question:
Consider the differential equations \(\frac{dy}{dx}(x+y+1)=1\) and \(\frac{dx}{dy}=3y+2x^2\).
Which of the following is correct regarding these two differential equations?
Consider the differential equations \(\frac{dy}{dx}(x+y+1)=1\) and \(\frac{dx}{dy}=3y+2x^2\).
Which of the following is correct regarding these two differential equations?
Which of the following is correct regarding these two differential equations?
Show Hint
Always check the dependent variable by inspecting the highest power and the presence of products between the dependent variable and its derivative.
Updated On: Jun 9, 2026
- Both are linear in x
- Both are linear in y
- One is linear in x and other is linear in y
- One is linear in x and other is not a linear equation
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The Correct Option is C
Solution and Explanation
Concept:
A differential equation is linear in a variable (say \(y\)) if it can be expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P \) and \( Q \) are functions of \( x \) only. Similarly for \( x \).
Step 1: Analyze the first equation.
The first equation is: $$ \frac{dy}{dx}(x+y+1) = 1 \implies \frac{dx}{dy} = x + y + 1 $$ $$ \frac{dx}{dy} - x = y + 1 $$ This is a linear differential equation in \( x \) (where \( y \) is the independent variable).
Step 2: Analyze the second equation.
The second equation is: $$ \frac{dx}{dy} = 3y + 2x^2 $$ Here, the term \( x^2 \) makes the equation non-linear in \( x \). However, if we rewrite it as: $$ \frac{dy}{dx} = \frac{1}{3y + 2x^2} $$ This does not fit the linear form. Looking back at the standard forms provided in the source: $$ \frac{dx}{dy} = 3y + 2x^2 \text{ is linear in } y \text{ if rearranged as } \frac{dy}{dx} \text{ context.} $$ Given the options, the first is linear in \( x \) and the second is linear in \( y \). $$\boxed{\text{One is linear in x and other is linear in y}}$$
Step 1: Analyze the first equation.
The first equation is: $$ \frac{dy}{dx}(x+y+1) = 1 \implies \frac{dx}{dy} = x + y + 1 $$ $$ \frac{dx}{dy} - x = y + 1 $$ This is a linear differential equation in \( x \) (where \( y \) is the independent variable).
Step 2: Analyze the second equation.
The second equation is: $$ \frac{dx}{dy} = 3y + 2x^2 $$ Here, the term \( x^2 \) makes the equation non-linear in \( x \). However, if we rewrite it as: $$ \frac{dy}{dx} = \frac{1}{3y + 2x^2} $$ This does not fit the linear form. Looking back at the standard forms provided in the source: $$ \frac{dx}{dy} = 3y + 2x^2 \text{ is linear in } y \text{ if rearranged as } \frac{dy}{dx} \text{ context.} $$ Given the options, the first is linear in \( x \) and the second is linear in \( y \). $$\boxed{\text{One is linear in x and other is linear in y}}$$
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