Given the differential equation: \[ \frac{y}{x^2}y' + e^{x^2}y^3 = 0 \] Let's find an **implicit solution** step by step. --- ### **Step 1: Rewrite \( y' \) as \( \frac{dy}{dx} \)** \[ \frac{y}{x^2} \frac{dy}{dx} + e^{x^2} y^3 = 0 \] --- ### **Step 2: Move the second term to the right** \[ \frac{y}{x^2} \frac{dy}{dx} = -e^{x^2} y^3 \] --- ### **Step 3: Divide both sides by \( y \) (assuming \( y \neq 0 \))** \[ \frac{1}{x^2} \frac{dy}{dx} = -e^{x^2} y^2 \] --- ### **Step 4: Separate the variables** \[ \frac{1}{y^2} dy = -x^2 e^{x^2} dx \] --- ### **Step 5: Integrate both sides** #### Left Side: \[ \int \frac{1}{y^2} dy = -\int x^2 e^{x^2} dx \] #### Left side integral: \[ \int y^{-2} dy = -y^{-1} = -\frac{1}{y} \] --- #### Right side integral (Let’s use substitution): Let \( u = x^2 \Rightarrow du = 2x dx \Rightarrow \frac{du}{2} = x dx \). But we have \( x^2 e^{x^2} dx \): Let’s differentiate by parts: Let \( u = x, dv = x e^{x^2} dx \). But an easier way is to recall: \[ \int x^n e^{a x^2} dx = \frac{e^{a x^2}}{2a} x^{n-1} + \text{(more terms if } n>1) \] But let's do it by parts: Let’s try substitution \( u = x^2 \Rightarrow du = 2x dx \rightarrow dx = \frac{du}{2x} \): So, \[ x^2 e^{x^2} dx = x^2 e^{u} \frac{du}{2x} = \frac{x e^{u}}{2} du \] But this seems to complicate things further. Alternatively, integrate by parts: Let \( u = x, dv = x e^{x^2} dx \). Or, let's directly differentiate \( e^{x^2} \): Let’s try integrating by parts: Let \( u = x, dv = x e^{x^2} dx \). But \( dv = x e^{x^2} dx \) and integrating \( dv \): Let’s set \( v = \int x e^{x^2} dx \). Let’s let \( w = x^2 \), \( dw = 2x dx \implies x dx = \frac{dw}{2} \): So, \[ \int x e^{x^2} dx = \int e^{w} \frac{dw}{2} = \frac{1}{2} e^{w} = \frac{1}{2} e^{x^2} \] Therefore, integrate by parts: Let \( u = x, dv = x e^{x^2} dx \) \[ du = dx \\ v = \frac{1}{2} e^{x^2} \] So, \[ \int x x e^{x^2} dx = \int x^2 e^{x^2} dx = x \cdot \frac{1}{2} e^{x^2} - \int \frac{1}{2} e^{x^2} dx \] So, \[ \int x^2 e^{x^2} dx = \frac{x}{2} e^{x^2} - \frac{1}{2} \int e^{x^2} dx \] --- ### **Step 6: Substitute back and write the implicit solution** Recall: \[ -\frac{1}{y} = -\left(\frac{x}{2} e^{x^2} - \frac{1}{2} \int e^{x^2} dx \right) + C \] Multiply both sides by -1 and rearrange: \[ \frac{1}{y} = \frac{x}{2} e^{x^2} - \frac{1}{2} \int e^{x^2} dx + C \] Or, combine constants: \[ \boxed{ \frac{1}{y} + \frac{1}{2} \int e^{x^2} dx = \frac{x}{2} e^{x^2} + C } \] This is the **implicit solution** to the given separable differential equation. --- **Final Answer (Implicit Solution):** \[ \boxed{ \frac{1}{y} + \frac{1}{2} \int e^{x^2} dx = \frac{x}{2} e^{x^2} + C } \]

Let's carefully analyze and solve the differential equation: \[ \frac{y}{x^2} y' + e^{2x^2} y^3 = 0 \] --- ### Step 1: Rewrite \( y' \) as \( \frac{dy}{dx} \) \[ \frac{y}{x^2} \frac{dy}{dx} + e^{2x^2} y^3 = 0 \] --- ### Step 2: Isolate \( \frac{dy}{dx} \) \[ \frac{y}{x^2} \frac{dy}{dx} = - e^{2x^2} y^3 \] Divide both sides by \( y \) (assuming \( y \neq 0 \)): \[ \frac{1}{x^2} \frac{dy}{dx} = - e^{2x^2} y^2 \] --- ### Step 3: Separate variables \[ \frac{1}{y^2} dy = - x^2 e^{2x^2} dx \] --- ### Step 4: Integrate both sides **Left side:** \[ \int y^{-2} dy = - \int x^2 e^{2x^2} dx \] Compute the left integral: \[ \int y^{-2} dy = - y^{-1} + C_1 \] **Right side:** \[ - \int x^2 e^{2x^2} dx \] --- ### Step 5: Focus on the right integral: \[ I = \int x^2 e^{2x^2} dx \] Use substitution: Let \( u = x^2 \Rightarrow du = 2x dx \Rightarrow x dx = \frac{du}{2} \). Express \( x^2 \) in terms of \( u \): \[ x^2 = u \] Rewrite \( I \): \[ I = \int u e^{2u} dx \] But \( dx = \frac{du}{2x} \). Since \( x = \sqrt{u} \), then: \[ dx = \frac{du}{2 \sqrt{u}} \] Substitute into the integral: \[ I = \int u e^{2u} \frac{du}{2 \sqrt{u}} = \frac{1}{2} \int u e^{2u} u^{-1/2} du = \frac{1}{2} \int u^{1 - 1/2} e^{2u} du = \frac{1}{2} \int u^{1/2} e^{2u} du \] So, \[ I = \frac{1}{2} \int u^{1/2} e^{2u} du \] --- ### Step 6: Solve \( J = \int u^{1/2} e^{2u} du \) This integral can be approached via **integration by parts**: Let: - \( w = u^{1/2} \Rightarrow dw = \frac{1}{2} u^{-1/2} du \) - \( dv = e^{2u} du \Rightarrow v = \frac{1}{2} e^{2u} \) Applying parts: \[ J = u^{1/2} \cdot \frac{1}{2} e^{2u} - \int \frac{1}{2} e^{2u} \cdot \frac{1}{2} u^{-1/2} du \] Simplify: \[ J = \frac{1}{2} u^{1/2} e^{2u} - \frac{1}{4} \int u^{-1/2} e^{2u} du \] Now, observe that: \[ I = \frac{1}{2} J = \frac{1}{2} \left[ \frac{1}{2} u^{1/2} e^{2u} - \frac{1}{4} \int u^{-1/2} e^{2u} du \right] \] Thus, \[ I = \frac{1}{4} u^{1/2} e^{2u} - \frac{1}{8} \int u^{-1/2} e^{2u} du \] Note that: \[ \int u^{-1/2} e^{2u} du \] is similar to the integral \( J \) but with the power of \( u \) being \( -1/2 \). Define: \[ K = \int u^{-1/2} e^{2u} du \] Use integration by parts again: - \( w = u^{-1/2} \Rightarrow dw = -\frac{1}{2} u^{-3/2} du \) - \( dv = e^{2u} du \Rightarrow v = \frac{1}{2} e^{2u} \) Applying parts: \[ K = u^{-1/2} \cdot \frac{1}{2} e^{2u} - \int \frac{1}{2} e^{2u} \cdot \left( - \frac{1}{2} u^{-3/2} \right) du \] Simplify: \[ K = \frac{1}{2} u^{-1/2} e^{2u} + \frac{1}{4} \int u^{-3/2} e^{2u} du \] The integral involves increasingly negative powers of \( u \). Since these integrals get complicated, a better approach is to **accept** that the indefinite integral involves special functions (like the exponential integral or incomplete gamma functions) and express the solution in terms of these. --- ### **Final expression:** In summary, the integral: \[ I = \int x^2 e^{2x^2} dx \] can be expressed as: \[ I = \frac{1}{4} u^{1/2} e^{2u} - \frac{1}{8} \int u^{-1/2} e^{2u} du \] with \( u = x^2 \). --- ### **Step 7: Implicit solution** Recall: \[ - y^{-1} = - I + C \] or: \[ \frac{1}{y} = I + C \] Substituting back: \[ \boxed{ \frac{1}{y} = \frac{1}{4} x e^{2x^2} - \frac{1}{8} \int_{0}^{x^2} t^{-1/2} e^{2t} dt + C } \] where we've replaced \( u = x^2 \), and the integral is in terms of \( t \). --- ### **Final answer:** \[ \boxed{ \frac{1}{y} = \frac{1}{4} x e^{2x^2} - \frac{1}{8} \int_{0}^{x^2} t^{-1/2} e^{2t} dt + C } \] This is the **implicit solution** to the differential equation: \[ \frac{y}{x^2} y' + e^{2x^2} y^3 = 0 \] --- **Note:** The integral involving \( t^{-1/2} e^{2t} dt \) is a non-elementary integral and can be expressed in terms of special functions (like the incomplete gamma function), but for most purposes, leaving it in integral form suffices.